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A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.
(Hint: The moment of inertia of the door about the vertical axis at one end is ML2/3.)
Mass of the bullet, m = 10 g = 10 × 10–3 kg
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:
As shown in the Figure, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m/s2)
(Hint: Consider the equilibrium of each side of the ladder separately.)
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.
A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m2.
(a) What is his new angular speed? (Neglect friction.)
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?
(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.
Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω1 and ω2 are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω1 ≠ ω2.
(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.
(a) Prove the theorem of perpendicular axes.
(Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin perpendicular to the plane is x2 + y2).
(b) Prove the theorem of parallel axes.
(Hint: If the centre of mass is chosen to be the origin 
).
(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–y plane 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 n particles, having masses m1, m2, m3, … , mn, at perpendicular distances r1, r2, r3, … , rn respectively 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,
Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by 
.
Using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
A body rolling on an inclined plane of height h,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.
A disc rotating about its axis with angular speed ωois placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. 7.41? Will the disc roll in the direction indicated?
Let be the angular velocity of the disc about the axis through O. The linear velocity of point A is,
Similarly, in the direction opposite to
(to the left).
Friction is necessary for the disc to roll, in case of a perfectly frictionless table, the disc will not roll.
Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins?
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.
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is μk = 0.2
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.
A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction µs = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination θof the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly
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:
Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
(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.
Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:
(a) Show pi = p’i + miV
Where pi is the momentum of the ith particle (of mass mi) and p′ i = miv′ i. Note v′ i is the velocity of the ith particle relative to the centre of mass.
Also, prove using the definition of the centre of mass 
(b) Show K = K′ + ½MV2
Where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and MV2/2 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14.
(c) Show L = L′ + R × MV
Where 
is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember r’i = ri – R; rest of the notation is the standard notation used in the chapter. Note L′ and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.
(d) Show 
Further, show that
where τ′ext is the sum of all external torques acting on the system about the centre of mass.
(Hint: Use the definition of centre of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)
(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:
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