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Chapter 7 System of particles and Rotational Motion Solutions

Question - 21 : - The oxygen molecule has a mass of 5.30 x 10–26 kg and a moment of inertia of 1.94 x10-46 kg m2 aboutan axis through its center perpendicular to the line joining the two atoms.Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kineticenergy of rotation is two thirds of its kinetic energy of translation. Find theaverage angular velocity of the molecule.

Answer - 21 : -

Here, m = 5.30 x 10–26 kg
I = 1.94 X 10-46 kg m2
υ = 500 m/s
If m/2 is mass of each atom of oxygen and 2r is distance between the two atomsas shown in Fig., then

Question - 22 : - A cylinder rolls up an inclined plane of angle ofinclination 30°. At the bottom of the inclined plane the center of mass of thecylinder has a speed of 5 m/s.
(a) How far will the cylinder go up the plane ?
(b) How long will it take to returif to the bottom ?

Answer - 22 : -

Here, θ = 30°, υ= 5 m/s
Let the cylinder go up the plane upto a height h.

Question - 23 : - As shown in Fig., the two sides of a step ladder BAand CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. Aweight 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.)

Answer - 23 : -

Let T be the tension in the rope DE. RB and Rc are the normalreactions of the floor at B and C
respectively. Since the ladder is in translational equilibrium, therefore, RB + Rc = W = mg = 40 x 9-8 =392 N …(i)
Ladder is also in rotational equilibrium, therefore, net torque on arms AB andAC is zero.
For arm AB, RB x BG– W x IG = T x AJ

Question - 24 : - A man stands on a rotating platform, with his armsstretched horizontally holding a 5 kg weight in each hand. The angular speed ofthe platform is 30 revolutions per minute. The man then brings his arms closeto his body with the distance of each weight from the axis changing from 90 cmto 20 cm. The moment of inertia of the man together with the platform may betaken to be constant and equal to 7.6 kg m2.What is his new angular speed ? (Neglect friction.)Is kinetic energy conservedin the process ? If not, from where does the change come about ?

Answer - 24 : -

No, kinetic energy is not conserved in the process. In fact, asmoment of inertia decreases, K.E. of rotation increases. This change comesabout as work is done by the man in bringing his arms closer to his body.

Question - 25 : - A bullet of mass 10 g and speed 500 m/s is fired intoa door and gets embedded exactly at the center of the door. The door is 1.0 mwide and weighs 12 kg. It is hinged at one end and rotates about a verticalaxis practically without friction. Find the angular speed of the door justafter the bullet embeds into it.(Hint. The moment of inertia of the door aboutthe vertical axis at one end is ML2/3.)

Answer - 25 : -

Question - 26 : - Two discs of moments of inertia I, and I2 about their respectiveaxes (normal to the disc and passing through the center), and rotating withangular speeds ω1 andω2 are brought intocontact 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 ofthe initial kinetic energies of the two discs. How do you account for this lossin energy ? Take ω1 ≠ω2

Answer - 26 : - Initial angular moment of the discs = I1ω1 + I2ω2
M.I. of two discs combined as a system = I1+ I2
Final angular moment of the combination = (I+ I2)ω
By using law of conservation of angular momentum,
we get I1ω1 + I2ω2 =( I1+I2 )ω

As the above term comes out to be positive, thus, rotationalkinetic energy of combined disc is less than the total initial energy.

Question - 27 : - (a) Prove the theorem of perpendicular axes (Hint.Square of the distance of a point (x, y) in the x-y plane from an axisperpendicular to the plane through the origin is x2 +y2).
(b) Prove the theorem of parallel axes (Hint. If the center of mass is chosento be the origin εmiri= 0).

Answer - 27 : - (a) The theorem ofperpendicular axes: According to thistheorem, the moment of inertia of a plane lamina (i.e., a two dimensional bodyof any shape/size) about any axis OZ perpendicular to the plane of the laminais equal to sum of the moments of inertia of the lamina about any two mutuallyperpendicular axes OX and OY in the plane of lamina, meeting at a point wherethe given axis OZ passes through the lamina. Suppose at the point ‘R’ m{particle is situated moment of inertia about Z axis of lamina
= moment of inertia of body about r-axis
= moment of inertia of body about y-axis.

(b) Theorem of parallelaxes: According to this theorem, moment of inertia ofa rigid body about
any axis AB is equal to moment of inertia of the body about another axis KLpassing through
centre of mass C of the body in a direction parallel to AB, plus the product oftotal mass M of the
body and square of the perpendicular distance between the two parallel axes. Ifh is perpendicular distance between the axes AB and KL, then Suppose rigid bodyis made up of n particles m1, m2, …. mn, mn at perpendicular distances r1, r2, ri….rn. respectively fromthe axis KL passing through centre of mass C of the body.
If h is the perpendicular distance of the particle of mass m{ from KL, then

Question - 28 : - Prove the result that the velocity of translation of arolling body (like a ring, disc, cylinder or sphere) at the bottom of aninclined plane of a height h is given by
using dynamical consideration (i.e., by considerationof forces and torques). Note k is the radius of gyration of the body about itssymmetry axis, and R is the radius of the body. The body starts from rest atthe top of the plane.

Answer - 28 : - When a body rolls down an incline of height h, weapply the principle of conservation of energy.
K.E. of translation + K.E. of rotation = P.E. at the top.

Question - 29 : - A disc rotating about its axis with angular speed ω0 is placed lightly(without any translational push) on a perfectly frictionless table. The radiusof the disc is R. What are the linear velocities of the points A, B and C onthe disc shown in Fig. ? Will the disc roll in the direction indicated ?

Answer - 29 : -

The disc will not roll in the given direction because frictionis necessary for the same.

Question - 30 : - Explain why friction is necessary to make the disc inFig. shown in Q. 28, roll in the direction indicated.
(a) Give the direction of frictional force at B, and the sense of frictionaltorque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins ?

Answer - 30 : -

To make the disc roll, torque is required. This torque will beprovided by the frictional force.
(a) Atpoint B, the frictional force supports the angular motion of this point, sofrictional force is in the direction of the arrow itself. Direction offrictional torque is normal to the paper in outward direction.
(b) Frictionalforce tries to decrease the velocity of the point B. When this velocity becomeszero, prefect rolling beings. For zero velocity, force of friction also becomeszero.

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