Ans: 1
Geometric centre; No
The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. For the given geometric shapes having a uniform mass density, the C.M. lies at their respective geometric centres.
The centre of mass of a body need not necessarily lie within it. For example, the C.M. of bodies such as a ring, a hollow sphere, etc., lies outside the body.
Ans: 2
The given situation can be shown as:
Distance between H and Cl atoms = 1.27Å
Mass of H atom = m
Mass of Cl atom = 35.5m
Let the centre of mass of the system lie at a distance xfrom the Cl atom.
Distance of the centre of mass from the H atom = (1.27 – x)
Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:
Here, the negative sign indicates that the centre of mass lies at the left of the molecule. Hence, the centre of mass of the HCl molecule lies 0.037Å from the Cl atom.
Ans: 3
No change
The child is running arbitrarily on a trolley moving with velocityv. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the boy’s motion is purely internal. Internal forces produce no effect on the motion of the bodies on which they act. Since no external force is involved in the boy–trolley system, the boy’s motion will produce no change in the velocity of the centre of mass of the trolley.
Ans: 4
Consider two vectors
and
, inclined at an angle θ,as shown in the following figure.
InΔOMN, we can write the relation:
= 2 × Area of ΔOMK
∴Area of ΔOMK
Ans: 5
Aparallelepiped with origin O and sides a,b, and cis shown in the following figure.
Volume of the given parallelepiped = abc
Let
be a unit vector perpendicular to both band c. Hence, and a have the same direction.
∴
=abc cosθ
=abc cos 0°
=abc
= Volume of the parallelepiped
Ans: 6
lx= ypz– zpy
ly= zpx– xpz
lz= xpy–ypx
Linear momentum of the particle,
Position vector of the particle,
Angular momentum,
Comparing the coefficients of
we get:
The particle moves in the x-yplane. Hence, the z-component of the position vector and linear momentum vector becomes zero, i.e.,
z= pz= 0
Thus, equation (i) reduces to:
Therefore, when the particle is confined to move in the x-yplane, the direction of angular momentum is along the z-direction.
Ans: 7
Let at a certain instant two particles be at points P and Q, as shown in the following figure.
Angular momentum of the system about point P:
Angular momentum of the system about point
Considera point R, which is at a distance yfrom point Q, i.e.,
QR =y
∴PR = d – y
Angular momentum of the system about point R:
Comparing equations (i), (ii), and (iii), we get:
We infer from equation (iv) that the angular momentum of a system does not depend on the point about which it is taken.
Ans: 8
The free body diagram of the bar is shown in the following figure.
Length of the bar, l= 2 m
T1and T2are the tensions produced in the left and right strings respectively.
Attranslational equilibrium, we have:
For rotational equilibrium, on taking the torque about the centre of gravity, we have:
Hence,the C.G. (centre of gravity) of the given bar lies 0.72 m from its left end.
Ans: 9
Mass of the car, m= 1800 kg
Distance between the front and back axles, d= 1.8 m
Distance between the C.G. (centre of gravity) and the back axle = 1.05 m
The various forces acting on the car are shown in the following figure.
Rfand Rbare the forces exerted by the level ground on the front and back wheels respectively.
At translational equilibrium:
=mg
= 1800 × 9.8
= 17640 N … (i)
For rotational equilibrium, on taking the torque about the C.G., we have:
Solving equations (i) and (ii), we get:
∴Rb= 17640 – 7350 = 10290 N
Therefore, the force exerted on each front wheel
, and
Theforce exerted on each back wheel
Ans: 10
(a)
The moment of inertia (M.I.) of a sphere about its diameter
According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
TheM.I. about a tangent of the sphere
(b)
The moment of inertia of a disc about its diameter =
According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.
TheM.I. of the disc about its centre
The situation is shown in the given figure.
Applyingthe theorem of parallel axes:
The moment of inertia about an axis normal to the disc and passing through a point on its edge
Ans: 11
Let m and r be the respective masses of the hollow cylinder and the solid sphere.
The moment of inertia of the hollow cylinder about its standard axis,
The moment of inertia of the solid sphere about an axis passing through its centre,
We have the relation:
Where,
α = Angular acceleration
τ = Torque
I = Moment of inertia
For the hollow cylinder,
For the solid sphere,
As an equal torque is applied to both the bodies,
Now, using the relation:
Where,
ω0 = Initial angular velocity
t = Time of rotation
ω = Final angular velocity
For equal ω0 and t, we have:
ω ∝ α … (ii)
From equations (i) and (ii), we can write:
ωII > ωI
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.
Ans: 12
Mass of the cylinder, m= 20 kg
Angular speed, ω= 100 rad s–1
Radius of the cylinder, r= 0.25 m
The moment of inertia of the solid cylinder:
∴Kinetic energy
∴Angular momentum, L= Iω
= 6.25 × 100
= 62.5 Js
Ans: 13
a) 100 rev/min
Initial angular velocity, ω1= 40 rev/min
Final angular velocity = ω2
The moment of inertia of the boy with stretched hands = I1
The moment of inertia of the boy with folded hands = I2
The two moments of inertia are related as:
Since no external force acts on the boy, the angular momentum Lis a constant.
Hence, for the two situations, we can write:
(b)Final K.E. = 2.5 Initial K.E.
Final kinetic rotation, EF
Initial kinetic rotation, EI
The increase in the rotational kinetic energy is attributed to the internal energy of the boy.
Ans: 14
Mass of the hollow cylinder, m= 3 kg
Radius of the hollow cylinder, r= 40 cm = 0.4 m
Applied force, F = 30 N
The moment of inertia of the hollow cylinder about its geometric axis:
I= mr2
= 3 × (0.4)2= 0.48 kg m2
Torque,
= 30 × 0.4 = 12 Nm
For angular acceleration
, torque is also given by the relation:
Linear acceleration = rα= 0.4 × 25 = 10 m s–2
Ans: 15
Angular speed of the rotor, ω =200 rad/s
Torque required, τ = 180 Nm
The power of the rotor (P) is related to torque and angular speed by the relation:
P = τω
= 180 × 200 = 36 × 103
= 36 kW
Hence, the power required by the engine is 36 kW.
Ans: 16
R/6; from the original centre of the body and opposite to the centre of the cut portion.
Mass per unit area of the original disc = σ
Radius of the original disc = R
Mass of the original disc, M = πR2σ
The disc with the cut portion is shown in the following figure:
Radius of the smaller disc =
Mass of the smaller disc, M’ =
Let O and O′ be the respective centres of the original disc and the disc cut off from the original. As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O′.
It is given that:
OO′=
After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are:
M (concentrated at O), and
–M′
concentrated at O′
(The negative sign indicates that this portion has been removed from the original disc.)
Let x be the distance through which the centre of mass of the remaining portion shifts from point O.
The relation between the centres of masses of two masses is given as:
For the given system, we can write:
(The negative sign indicates that the centre of mass gets shifted toward the left of point O.)
Ans: 17
LetW and W′be the respective weights of the metre stick and the coin.
The mass of the metre stick is concentrated at its mid-point, i.e., at the 50 cm mark.
Mass of the meter stick = m’
Mass of each coin, m= 5 g
When the coins are placed 12 cm away from the end P, the centre of mass gets shifted by 5 cm from point R toward the end P. The centre of mass is located at a distance of 45 cm from point P.
The net torque will be conserved for rotational equilibrium about point R.
Hence, the mass of the metre stick is 66 g.
Ans: 18
Answer:(a)Yes (b)Yes (c) On the smaller inclination
(a)Mass of the sphere = m
Height of the plane = h
Velocity of the sphere at the bottom of the plane =v
At the top of the plane, the total energy of the sphere = Potential energy = mgh
At the bottom of the plane, the sphere has both translational and rotational kinetic energies.
Hence, total energy =
Usingthe law of conservation of energy, we can write:
For a solid sphere, the moment of inertia about its centre,
Hence, equation (i) becomes:
Hence, the velocity of the sphere at the bottom depends only on height (h) and acceleration due to gravity (g). Both these values are constants. Therefore, the velocity at the bottom remains the same from whichever inclined plane the sphere is rolled.
(b), (c)Consider two inclined planes with inclinations θ1and θ2, related as:
θ1< θ2
The acceleration produced in the sphere when it rolls down the plane inclined at θ1is:
g sin θ1
The various forces acting on the sphere are shown in the following figure.
R1is the normal reaction to the sphere.
Similarly, the acceleration produced in the sphere when it rolls down the plane inclined at θ2is:
g sin θ2
The various forces acting on the sphere are shown in the following figure.
R2is the normal reaction to the sphere.
θ2> θ1; sin θ2> sin θ1 ... (i)
∴a2> a1 … (ii)
Initial velocity, u = 0
Final velocity, v= Constant
Using the first equation of motion, we can obtain the time of roll as:
v= u + at
From equations (ii) and (iii), we get:
t2< t1
Hence, the sphere will take a longer time to reach the bottom of the inclined plane having the smaller inclination.
Ans: 19
Radius of the hoop, r= 2 m
Mass of the hoop, m= 100 kg
Velocity of the hoop, v= 20 cm/s = 0.2 m/s
Total energy of the hoop = Translational KE + Rotational KE
Moment of inertia of the hoop about its centre, I= mr2
The work required to be done for stopping the hoop is equal to the total energy of the hoop.
∴Required work to be done, W= mv2= 100 × (0.2)2= 4 J
Ans: 20
Mass of an oxygen molecule, m= 5.30 × 10–26kg
Moment of inertia, I= 1.94 × 10–46kg m2
Velocity of the oxygen molecule, v= 500 m/s
The separation between the two atoms of the oxygen molecule = 2r
Mass of each oxygen atom =
Hence, moment of inertia I, is calculated as:
It is given that:
Ans: 21
A solid cylinder rolling up an inclination is shown in the following figure.
Initial velocity of the solid cylinder, v = 5 m/s
Angle of inclination, θ = 30°
Height reached by the cylinder = h
(a) Energy of the cylinder at point A:
Energy of the cylinder at point B = mgh
Using the law of conservation of energy, we can write:
Moment of inertia of the solid cylinder,
In ΔABC:
Hence, the cylinder will travel 3.82 m up the inclined plane.
(b) For radius of gyration K, the velocity of the cylinder at the instance when it rolls back to the bottom is given by the relation:
The time taken to return to the bottom is:
Therefore, the total time taken by the cylinder to return to the bottom is (2 × 0.764) 1.53 s.
Ans: 22
The given situation can be shown as:
NB = Force exerted on the ladder by the floor point B
NC = Force exerted on the ladder by the floor point C
T = Tension in the rope
BA = CA = 1.6 m
DE = 0. 5 m
BF = 1.2 m
Mass of the weight, m = 40 kg
Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H.
ΔABI and ΔAIC are similar
∴BI = IC
Hence, I is the mid-point of BC.
DE || BC
BC = 2 × DE = 1 m
AF = BA – BF = 0.4 m … (i)
D is the mid-point of AB.
Hence, we can write:
Using equations (i) and (ii), we get:
FE = 0.4 m
Hence, F is the mid-point of AD.
FG||DH and F is the mid-point of AD. Hence, G will also be the mid-point of AH.
ΔAFG and ΔADH are similar
In ΔADH:
For translational equilibrium of the ladder, the upward force should be equal to the downward force.
Nc + NB = mg = 392 … (iii)
For rotational equilibrium of the ladder, the net moment about A is:
Adding equations (iii) and (iv), we get:
For rotational equilibrium of the side AB, consider the moment about A.
Ans: 23
(a)58.88 rev/min(b) No
(a)Moment of inertia of the man-platform system = 7.6 kg m2
Moment of inertia when the man stretches his hands to a distance of 90 cm:
2×m r2
= 2 ×5 ×(0.9)2
= 8.1 kg m2
Initial moment of inertia of the system,
Angular speed,
Angular momentum,
Moment of inertia when the man folds his hands to a distance of 20 cm:
2×mr2
= 2 ×5 (0.2)2= 0.4 kg m2
Final moment of inertia,
Final angular speed =
Final angular momentum,
… (ii)
From the conservation of angular momentum, we have:
(b)Kinetic energy is not conserved in the given process. In fact, with the decrease in the moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands toward himself.
Ans: 24
Mass of the bullet, m= 10 g = 10 ×10–3kg
Velocity of the bullet, v= 500 m/s
Thickness of the door, L= 1 m
Radius of the door,r=
Mass of the door, M= 12 kg
Angular momentum imparted by the bullet on the door:
α=mvr
Moment of inertia of the door:
Ans: 25
(a)
When the two discs are joined together, their moments of inertia get added up.
Moment of inertia of the system of two discs,
Let ω be the angular speed of the system.
Total final angular momentum,
Using the law of conservation of angular momentum, we have:
(b)Kinetic energy of disc I,
Kinetic energy of disc II,
Total initial kinetic energy,
When the discs are joined, their moments of inertia get added up.
Moment of inertia of the system,
Angular speed of the system = ω
Final kinetic energy Ef:
The loss of KE can be attributed to the frictional force that comes into play when the two discs come in contact with each other.
Ans: 26
(a)The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.
A physical body with centre O and a point mass m,in the x–yplane at (x,y) is shown in the following figure.
Moment of inertia about x-axis,Ix= mx2
Moment of inertia about y-axis,Iy= my2
Moment of inertia about z-axis,Iz=
Ix+ Iy= mx2+ my2
= m(x2+ y2)
(b)The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
Suppose a rigid body is made up of nparticles, having masses m1,m2,m3,… , mn, at perpendicular distances r1,r2,r3,… , rnrespectively from the centre of mass O of the rigid body.
The moment of inertia about axis RS passing through the point O:
IRS=
The perpendicular distance of mass mi, from the axis QP = a+ ri
Hence,the moment of inertia about axis QP:
Now, at the centre of mass, the moment of inertia of all the particles about the axis passing through the centre of mass is zero, that is,
Ans: 27
A body rolling on an inclined plane of heighth,is shown in the following figure:
m= Mass of the body
R= Radius of the body
K= Radius of gyration of the body
v= Translational velocity of the body
h=Height of the inclined plane
g = Acceleration due to gravity
Total energy at the top of the plane, E1=mgh
Total energy at the bottom of the plane,
But
From the law of conservation of energy, we have:
Hence, the given result is proved.
Ans: 28
vA = Rωo;vB = Rωo;
; The disc will not roll
Angular speed of the disc = ωo
Radius of the disc = R
Using the relation for linear velocity, v = ωoR
For point A:
vA = Rωo; in the direction tangential to the right
For point B:
vB = Rωo; in the direction tangential to the left
For point C:
in the direction same as that of vA
The directions of motion of points A, B, and C on the disc are shown in the following figure
Since the disc is placed on a frictionless table, it will not roll. This is because the presence of friction is essential for the rolling of a body.
Ans: 29
A torque is required to roll the given disc. As per the definition of torque, the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is required for making the disc roll.
(a) Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward. The sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.
(b) Since frictional force acts opposite to the direction of velocity at point B, perfect rolling will begin when the velocity at that point becomes equal to zero. This will make the frictional force acting on the disc zero.
Ans: 30
Disc
Radii of the ring and the disc, r = 10 cm = 0.1 m
Initial angular speed, ω0=10 π rad s–1
Coefficient of kinetic friction, μk = 0.2
Initial velocity of both the objects, u = 0
Motion of the two objects is caused by frictional force. As per Newton’s second law of motion, we have frictional force, f= ma
μkmg= ma
Where,
a = Acceleration produced in the objects
m = Mass
∴a = μkg … (i)
As per the first equation of motion, the final velocity of the objects can be obtained as:
v = u + at
= 0 + μkgt
= μkgt … (ii)
The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.
Torque, τ= –Iα
α = Angular acceleration
μxmgr = –Iα
Using the first equation of rotational motion to obtain the final angular speed:
Rolling starts when linear velocity, v = rω
Equating equations (ii) and (v), we get:
Since td > tr, the disc will start rolling before the ring.
Ans: 31
Mass of the cylinder, m= 10 kg
Radius of the cylinder, r= 15 cm = 0.15 m
Co-efficient of kinetic friction, µk= 0.25
Angle of inclination, θ= 30°
Moment of inertia of a solid cylinder about its geometric axis,
The various forces acting on the cylinder are shown in the following figure:
The acceleration of the cylinder is given as:
(a) Using Newton’s second law of motion, we can write net force as:
fnet= ma
(b) During rolling, the instantaneous point of contact with the plane comes to rest. Hence, the work done against frictional force is zero.
(c) For rolling without skid, we have the relation:
Ans: 32
(a) False
Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.
(b) True
Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero.
(c) False
When a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.
(d) True
When perfect rolling begins, the frictional force acting at the lowermost point becomes zero. Hence, the work done against friction is also zero.
(e) True
The rolling of a body occurs when a frictional force acts between the body and the surface. This frictional force provides the torque necessary for rolling. In the absence of a frictional force, the body slips from the inclined plane under the effect of its own weight.
Ans: 33
(a)Take a system of i moving particles.
Mass of the ith particle = mi
Velocity of the ith particle = vi
Hence, momentum of the ith particle, pi = mi vi
Velocity of the centre of mass = V
The velocity of the ith particle with respect to the centre of mass of the system is given as:
v’i = vi – V … (1)
Multiplying mi throughout equation (1), we get:
mi v’i = mi vi – mi V
p’i = pi – mi V
Where,
pi’ = mivi’ = Momentum of the ith particle with respect to the centre of mass of the system
∴pi = p’i + mi V
We have the relation: p’i = mivi’
Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we get:
(b) We have the relation for velocity of the ith particle as:
vi = v’i + V
… (2)
Taking the dot product of equation (2) with itself, we get:
Where,
K =
= Total kinetic energy of the system of particles
K’ =
= Total kinetic energy of the system of particles with respect to the centre of mass
= Kinetic energy of the translation of the system as a whole
(c) Position vector of the ith particle with respect to origin = ri
Position vector of the ith particle with respect to the centre of mass = r’i
Position vector of the centre of mass with respect to the origin = R
It is given that:
r’i = ri – R
ri = r’i + R
We have from part (a),
pi = p’i + mi V
Taking the cross product of this relation by ri, we get:
(d) We have the relation:
We have the relation:
Geometric centre; No
The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. For the given geometric shapes having a uniform mass density, the C.M. lies at their respective geometric centres.
The centre of mass of a body need not necessarily lie within it. For example, the C.M. of bodies such as a ring, a hollow sphere, etc., lies outside the body.
Ans: 2
The given situation can be shown as:
Distance between H and Cl atoms = 1.27Å
Mass of H atom = m
Mass of Cl atom = 35.5m
Let the centre of mass of the system lie at a distance xfrom the Cl atom.
Distance of the centre of mass from the H atom = (1.27 – x)
Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:
Here, the negative sign indicates that the centre of mass lies at the left of the molecule. Hence, the centre of mass of the HCl molecule lies 0.037Å from the Cl atom.
Ans: 3
No change
The child is running arbitrarily on a trolley moving with velocityv. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the boy’s motion is purely internal. Internal forces produce no effect on the motion of the bodies on which they act. Since no external force is involved in the boy–trolley system, the boy’s motion will produce no change in the velocity of the centre of mass of the trolley.
Ans: 4
Consider two vectors
and, inclined at an angle θ,as shown in the following figure.InΔOMN, we can write the relation:
= 2 × Area of ΔOMK
∴Area of ΔOMK
Ans: 5
Aparallelepiped with origin O and sides a,b, and cis shown in the following figure.
Volume of the given parallelepiped = abc
Let
be a unit vector perpendicular to both band c. Hence, and a have the same direction.∴
=abc cosθ
=abc cos 0°
=abc
= Volume of the parallelepiped
Ans: 6
lx= ypz– zpy
ly= zpx– xpz
lz= xpy–ypx
Linear momentum of the particle,
Position vector of the particle,
Angular momentum,
Comparing the coefficients of
we get:The particle moves in the x-yplane. Hence, the z-component of the position vector and linear momentum vector becomes zero, i.e.,
z= pz= 0
Thus, equation (i) reduces to:
Therefore, when the particle is confined to move in the x-yplane, the direction of angular momentum is along the z-direction.
Ans: 7
Let at a certain instant two particles be at points P and Q, as shown in the following figure.
Angular momentum of the system about point P:
Angular momentum of the system about point
Considera point R, which is at a distance yfrom point Q, i.e.,
QR =y
∴PR = d – y
Angular momentum of the system about point R:
Comparing equations (i), (ii), and (iii), we get:
We infer from equation (iv) that the angular momentum of a system does not depend on the point about which it is taken.
Ans: 8
The free body diagram of the bar is shown in the following figure.
Length of the bar, l= 2 m
T1and T2are the tensions produced in the left and right strings respectively.
Attranslational equilibrium, we have:
For rotational equilibrium, on taking the torque about the centre of gravity, we have:
Hence,the C.G. (centre of gravity) of the given bar lies 0.72 m from its left end.
Ans: 9
Mass of the car, m= 1800 kg
Distance between the front and back axles, d= 1.8 m
Distance between the C.G. (centre of gravity) and the back axle = 1.05 m
The various forces acting on the car are shown in the following figure.
Rfand Rbare the forces exerted by the level ground on the front and back wheels respectively.
At translational equilibrium:
=mg= 1800 × 9.8
= 17640 N … (i)
For rotational equilibrium, on taking the torque about the C.G., we have:
Solving equations (i) and (ii), we get:
∴Rb= 17640 – 7350 = 10290 N
Therefore, the force exerted on each front wheel
, andTheforce exerted on each back wheel
Ans: 10
(a)
The moment of inertia (M.I.) of a sphere about its diameter
According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
TheM.I. about a tangent of the sphere
(b)
The moment of inertia of a disc about its diameter =
According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.
TheM.I. of the disc about its centre
The situation is shown in the given figure.
Applyingthe theorem of parallel axes:
The moment of inertia about an axis normal to the disc and passing through a point on its edge
Ans: 11
Let m and r be the respective masses of the hollow cylinder and the solid sphere.
The moment of inertia of the hollow cylinder about its standard axis,
The moment of inertia of the solid sphere about an axis passing through its centre,
We have the relation:
Where,
α = Angular acceleration
τ = Torque
I = Moment of inertia
For the hollow cylinder,
For the solid sphere,
As an equal torque is applied to both the bodies,
Now, using the relation:
Where,
ω0 = Initial angular velocity
t = Time of rotation
ω = Final angular velocity
For equal ω0 and t, we have:
ω ∝ α … (ii)
From equations (i) and (ii), we can write:
ωII > ωI
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.
Ans: 12
Mass of the cylinder, m= 20 kg
Angular speed, ω= 100 rad s–1
Radius of the cylinder, r= 0.25 m
The moment of inertia of the solid cylinder:
∴Kinetic energy
∴Angular momentum, L= Iω
= 6.25 × 100
= 62.5 Js
Ans: 13
a) 100 rev/min
Initial angular velocity, ω1= 40 rev/min
Final angular velocity = ω2
The moment of inertia of the boy with stretched hands = I1
The moment of inertia of the boy with folded hands = I2
The two moments of inertia are related as:
Since no external force acts on the boy, the angular momentum Lis a constant.
Hence, for the two situations, we can write:
(b)Final K.E. = 2.5 Initial K.E.
Final kinetic rotation, EF
Initial kinetic rotation, EI
The increase in the rotational kinetic energy is attributed to the internal energy of the boy.
Ans: 14
Mass of the hollow cylinder, m= 3 kg
Radius of the hollow cylinder, r= 40 cm = 0.4 m
Applied force, F = 30 N
The moment of inertia of the hollow cylinder about its geometric axis:
I= mr2
= 3 × (0.4)2= 0.48 kg m2
Torque,
= 30 × 0.4 = 12 Nm
For angular acceleration
, torque is also given by the relation:Linear acceleration = rα= 0.4 × 25 = 10 m s–2
Ans: 15
Angular speed of the rotor, ω =200 rad/s
Torque required, τ = 180 Nm
The power of the rotor (P) is related to torque and angular speed by the relation:
P = τω
= 180 × 200 = 36 × 103
= 36 kW
Hence, the power required by the engine is 36 kW.
Ans: 16
R/6; from the original centre of the body and opposite to the centre of the cut portion.
Mass per unit area of the original disc = σ
Radius of the original disc = R
Mass of the original disc, M = πR2σ
The disc with the cut portion is shown in the following figure:
Radius of the smaller disc =
Mass of the smaller disc, M’ =
Let O and O′ be the respective centres of the original disc and the disc cut off from the original. As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O′.
It is given that:
OO′=
After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are:
M (concentrated at O), and
–M′
concentrated at O′(The negative sign indicates that this portion has been removed from the original disc.)
Let x be the distance through which the centre of mass of the remaining portion shifts from point O.
The relation between the centres of masses of two masses is given as:
For the given system, we can write:
(The negative sign indicates that the centre of mass gets shifted toward the left of point O.)
Ans: 17
LetW and W′be the respective weights of the metre stick and the coin.
The mass of the metre stick is concentrated at its mid-point, i.e., at the 50 cm mark.
Mass of the meter stick = m’
Mass of each coin, m= 5 g
When the coins are placed 12 cm away from the end P, the centre of mass gets shifted by 5 cm from point R toward the end P. The centre of mass is located at a distance of 45 cm from point P.
The net torque will be conserved for rotational equilibrium about point R.
Hence, the mass of the metre stick is 66 g.
Ans: 18
Answer:(a)Yes (b)Yes (c) On the smaller inclination
(a)Mass of the sphere = m
Height of the plane = h
Velocity of the sphere at the bottom of the plane =v
At the top of the plane, the total energy of the sphere = Potential energy = mgh
At the bottom of the plane, the sphere has both translational and rotational kinetic energies.
Hence, total energy =
Usingthe law of conservation of energy, we can write:
For a solid sphere, the moment of inertia about its centre,
Hence, equation (i) becomes:
Hence, the velocity of the sphere at the bottom depends only on height (h) and acceleration due to gravity (g). Both these values are constants. Therefore, the velocity at the bottom remains the same from whichever inclined plane the sphere is rolled.
(b), (c)Consider two inclined planes with inclinations θ1and θ2, related as:
θ1< θ2
The acceleration produced in the sphere when it rolls down the plane inclined at θ1is:
g sin θ1
The various forces acting on the sphere are shown in the following figure.
R1is the normal reaction to the sphere.
Similarly, the acceleration produced in the sphere when it rolls down the plane inclined at θ2is:
g sin θ2
The various forces acting on the sphere are shown in the following figure.
R2is the normal reaction to the sphere.
θ2> θ1; sin θ2> sin θ1 ... (i)
∴a2> a1 … (ii)
Initial velocity, u = 0
Final velocity, v= Constant
Using the first equation of motion, we can obtain the time of roll as:
v= u + at
From equations (ii) and (iii), we get:
t2< t1
Hence, the sphere will take a longer time to reach the bottom of the inclined plane having the smaller inclination.
Ans: 19
Radius of the hoop, r= 2 m
Mass of the hoop, m= 100 kg
Velocity of the hoop, v= 20 cm/s = 0.2 m/s
Total energy of the hoop = Translational KE + Rotational KE
Moment of inertia of the hoop about its centre, I= mr2
The work required to be done for stopping the hoop is equal to the total energy of the hoop.
∴Required work to be done, W= mv2= 100 × (0.2)2= 4 J
Ans: 20
Mass of an oxygen molecule, m= 5.30 × 10–26kg
Moment of inertia, I= 1.94 × 10–46kg m2
Velocity of the oxygen molecule, v= 500 m/s
The separation between the two atoms of the oxygen molecule = 2r
Mass of each oxygen atom =
Hence, moment of inertia I, is calculated as:
It is given that:
Ans: 21
A solid cylinder rolling up an inclination is shown in the following figure.
Initial velocity of the solid cylinder, v = 5 m/s
Angle of inclination, θ = 30°
Height reached by the cylinder = h
(a) Energy of the cylinder at point A:
Energy of the cylinder at point B = mgh
Using the law of conservation of energy, we can write:
Moment of inertia of the solid cylinder,
In ΔABC:
Hence, the cylinder will travel 3.82 m up the inclined plane.
(b) For radius of gyration K, the velocity of the cylinder at the instance when it rolls back to the bottom is given by the relation:
The time taken to return to the bottom is:
Therefore, the total time taken by the cylinder to return to the bottom is (2 × 0.764) 1.53 s.
Ans: 22
The given situation can be shown as:
NB = Force exerted on the ladder by the floor point B
NC = Force exerted on the ladder by the floor point C
T = Tension in the rope
BA = CA = 1.6 m
DE = 0. 5 m
BF = 1.2 m
Mass of the weight, m = 40 kg
Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H.
ΔABI and ΔAIC are similar
∴BI = IC
Hence, I is the mid-point of BC.
DE || BC
BC = 2 × DE = 1 m
AF = BA – BF = 0.4 m … (i)
D is the mid-point of AB.
Hence, we can write:
Using equations (i) and (ii), we get:
FE = 0.4 m
Hence, F is the mid-point of AD.
FG||DH and F is the mid-point of AD. Hence, G will also be the mid-point of AH.
ΔAFG and ΔADH are similar
In ΔADH:
For translational equilibrium of the ladder, the upward force should be equal to the downward force.
Nc + NB = mg = 392 … (iii)
For rotational equilibrium of the ladder, the net moment about A is:
Adding equations (iii) and (iv), we get:
For rotational equilibrium of the side AB, consider the moment about A.
Ans: 23
(a)58.88 rev/min(b) No
(a)Moment of inertia of the man-platform system = 7.6 kg m2
Moment of inertia when the man stretches his hands to a distance of 90 cm:
2×m r2
= 2 ×5 ×(0.9)2
= 8.1 kg m2
Initial moment of inertia of the system,
Angular speed,
Angular momentum,
Moment of inertia when the man folds his hands to a distance of 20 cm:
2×mr2
= 2 ×5 (0.2)2= 0.4 kg m2
Final moment of inertia,
Final angular speed =
Final angular momentum,
… (ii)From the conservation of angular momentum, we have:
(b)Kinetic energy is not conserved in the given process. In fact, with the decrease in the moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands toward himself.
Ans: 24
Mass of the bullet, m= 10 g = 10 ×10–3kg
Velocity of the bullet, v= 500 m/s
Thickness of the door, L= 1 m
Radius of the door,r=
Mass of the door, M= 12 kg
Angular momentum imparted by the bullet on the door:
α=mvr
Moment of inertia of the door:
Ans: 25
(a)
When the two discs are joined together, their moments of inertia get added up.
Moment of inertia of the system of two discs,
Let ω be the angular speed of the system.
Total final angular momentum,
Using the law of conservation of angular momentum, we have:
(b)Kinetic energy of disc I,
Kinetic energy of disc II,
Total initial kinetic energy,
When the discs are joined, their moments of inertia get added up.
Moment of inertia of the system,
Angular speed of the system = ω
Final kinetic energy Ef:
The loss of KE can be attributed to the frictional force that comes into play when the two discs come in contact with each other.
Ans: 26
(a)The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.
A physical body with centre O and a point mass m,in the x–yplane at (x,y) is shown in the following figure.
Moment of inertia about x-axis,Ix= mx2
Moment of inertia about y-axis,Iy= my2
Moment of inertia about z-axis,Iz=
Ix+ Iy= mx2+ my2
= m(x2+ y2)
(b)The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
Suppose a rigid body is made up of nparticles, having masses m1,m2,m3,… , mn, at perpendicular distances r1,r2,r3,… , rnrespectively from the centre of mass O of the rigid body.
The moment of inertia about axis RS passing through the point O:
IRS=
The perpendicular distance of mass mi, from the axis QP = a+ ri
Hence,the moment of inertia about axis QP:
Now, at the centre of mass, the moment of inertia of all the particles about the axis passing through the centre of mass is zero, that is,
Ans: 27
A body rolling on an inclined plane of heighth,is shown in the following figure:
m= Mass of the body
R= Radius of the body
K= Radius of gyration of the body
v= Translational velocity of the body
h=Height of the inclined plane
g = Acceleration due to gravity
Total energy at the top of the plane, E1=mgh
Total energy at the bottom of the plane,
But
From the law of conservation of energy, we have:
Hence, the given result is proved.
Ans: 28
vA = Rωo;vB = Rωo;
; The disc will not rollAngular speed of the disc = ωo
Radius of the disc = R
Using the relation for linear velocity, v = ωoR
For point A:
vA = Rωo; in the direction tangential to the right
For point B:
vB = Rωo; in the direction tangential to the left
For point C:
in the direction same as that of vAThe directions of motion of points A, B, and C on the disc are shown in the following figure
Since the disc is placed on a frictionless table, it will not roll. This is because the presence of friction is essential for the rolling of a body.
Ans: 29
A torque is required to roll the given disc. As per the definition of torque, the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is required for making the disc roll.
(a) Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward. The sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.
(b) Since frictional force acts opposite to the direction of velocity at point B, perfect rolling will begin when the velocity at that point becomes equal to zero. This will make the frictional force acting on the disc zero.
Ans: 30
Disc
Radii of the ring and the disc, r = 10 cm = 0.1 m
Initial angular speed, ω0=10 π rad s–1
Coefficient of kinetic friction, μk = 0.2
Initial velocity of both the objects, u = 0
Motion of the two objects is caused by frictional force. As per Newton’s second law of motion, we have frictional force, f= ma
μkmg= ma
Where,
a = Acceleration produced in the objects
m = Mass
∴a = μkg … (i)
As per the first equation of motion, the final velocity of the objects can be obtained as:
v = u + at
= 0 + μkgt
= μkgt … (ii)
The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.
Torque, τ= –Iα
α = Angular acceleration
μxmgr = –Iα
Using the first equation of rotational motion to obtain the final angular speed:
Rolling starts when linear velocity, v = rω
Equating equations (ii) and (v), we get:
Since td > tr, the disc will start rolling before the ring.
Ans: 31
Mass of the cylinder, m= 10 kg
Radius of the cylinder, r= 15 cm = 0.15 m
Co-efficient of kinetic friction, µk= 0.25
Angle of inclination, θ= 30°
Moment of inertia of a solid cylinder about its geometric axis,
The various forces acting on the cylinder are shown in the following figure:
The acceleration of the cylinder is given as:
(a) Using Newton’s second law of motion, we can write net force as:
fnet= ma
(b) During rolling, the instantaneous point of contact with the plane comes to rest. Hence, the work done against frictional force is zero.
(c) For rolling without skid, we have the relation:
Ans: 32
(a) False
Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.
(b) True
Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero.
(c) False
When a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.
(d) True
When perfect rolling begins, the frictional force acting at the lowermost point becomes zero. Hence, the work done against friction is also zero.
(e) True
The rolling of a body occurs when a frictional force acts between the body and the surface. This frictional force provides the torque necessary for rolling. In the absence of a frictional force, the body slips from the inclined plane under the effect of its own weight.
Ans: 33
(a)Take a system of i moving particles.
Mass of the ith particle = mi
Velocity of the ith particle = vi
Hence, momentum of the ith particle, pi = mi vi
Velocity of the centre of mass = V
The velocity of the ith particle with respect to the centre of mass of the system is given as:
v’i = vi – V … (1)
Multiplying mi throughout equation (1), we get:
mi v’i = mi vi – mi V
p’i = pi – mi V
Where,
pi’ = mivi’ = Momentum of the ith particle with respect to the centre of mass of the system
∴pi = p’i + mi V
We have the relation: p’i = mivi’
Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we get:
(b) We have the relation for velocity of the ith particle as:
vi = v’i + V
… (2)Taking the dot product of equation (2) with itself, we get:
Where,
K =
= Total kinetic energy of the system of particlesK’ =
= Total kinetic energy of the system of particles with respect to the centre of mass = Kinetic energy of the translation of the system as a whole(c) Position vector of the ith particle with respect to origin = ri
Position vector of the ith particle with respect to the centre of mass = r’i
Position vector of the centre of mass with respect to the origin = R
It is given that:
r’i = ri – R
ri = r’i + R
We have from part (a),
pi = p’i + mi V
Taking the cross product of this relation by ri, we get:
(d) We have the relation:
We have the relation:
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