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Chapter 7 System of particles and Rotational Motion Solutions

Question - 11 : - (a) Find the moment of inertia of a sphere about atangent to the sphere, given the moment of inertia of the sphere about any ofits diameters to be 2 MR2/5, where M isthe mass of the sphere and R is the radius of the sphere.
(b) What is the moment of inertia of a uniform disc of radius R and mass Mabout an axis
(1) passing through its center and normal to the disc
(2) passing through a point on its edge and normal to the disc ? The moment ofinertia of the disc about any of its diameters is given to be 1 MR2 .

Answer - 11 : -

Question - 12 : - Torques of equal magnitude are applied to a hollowcylinder and a solid sphere, both having the same mass and radius. The cylinderis free to rotate about its standard axis of symmetry, and the sphere is freeto rotate about an axis passing through its center. Which of the two willacquire a greater angular speed after a given time ?

Answer - 12 : - M.I. of the cylinder =I1= MR2
M.I. of the cylinder=I2=2 MR2




ω = ω0 +αt, therefore sphereacquires a greater speed than cylinder as α2 >α1

Question - 13 : - A stolid cylinder of mass 20 kg rotates about its axiswith angular speed 100 rad s-1. Theradius of the cylinder is 0.25 What is the kinetic energy associated with therotation of the cylinder ? What is the magnitude of angular momlntum of thecylinder about its axis ?

Answer - 13 : -

Question - 14 : - (a) Achild stands at the center of a turntable with his two arms outstretched. Theturntable is set rotating with an angular speed of 40 rev/min. How much is theangular speed of the child if he folds his hands back and thereby reduces hismoment of inertia to 2/5 times the initial value ? Assume that the turntablerotates without friction.
(b) Showthat the child’s new kinetic energy of rotation is more than the initialkinetic energy of rotation. How do you account for this increase in kineticenergy ?

Answer - 14 : - (a) Suppose,initial moment of inertia of the child is I1. Then final moment of inertia of inertia,
I2 = 2Also, v1= 40 rev min

Clearly,final (K.E.)rot becomesmore because the child uses his internal energy when he folds his hands toincrease the kinetic energy.

Question - 15 : - A rope of negligible mass is wound round a hollowcylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of thecylinder if the rope is pulled with a force of 30 N ? What is the linearacceleration of the rope ? Assume that there is no slipping.

Answer - 15 : - Here, M = 3 kg, R = 40 cm = 0.4 m
M.I. of the hollow cylinder about its axis = I = MR2 =3 x (0.4)2 = 0.48 kg m2
When force of 30 N is applied over the rope wound on the cylinder, thetorque will act on the cylinder. It is given by
τ= FR = 30 x 0.4 = 12 N m
If a is angular acceleration produced, then
τ = Iα

Question - 16 : - To maintain a rotor at a uniform angular speed of 200rad s-1, an engine needs to transmit atorque of 180 N m. What is the power required by the engine ? (Note : uniformangular velocity in the absence of friction implies zero torque. In practice,applied torque is needed to counter frictional torque). Assume that the engineis 100% efficient.

Answer - 16 : -

Here τ = 18 Nm, ω = 200 rad s-1
P =τω
∴ V P =180 x 200 = 36000 W = 36 kW.

Question - 17 : - From a uniform disk of radius R, a circular section ofradius R/2 is cut out. The center of the hole is at R/2 from the center of theoriginal disc. Locate the center of mass of the resulting flat body.

Answer - 17 : - Let mass of disc = M


Mass of the portion removed from disc is concentratedat O] and the mass of the remaining disc is
supposed to be concentrated at O2 at a distance x from the center of the disc (O).

Question - 18 : - A meter stick is balanced on a knife edge at itscenter. When two coins, each of mass 5 g are put one on top of the other at the12*0 cm mark, the stick is found to be balanced at 45-0 cm. What is the mass ofthe meter stick ?

Answer - 18 : -

Let m be the mass of the stick concentrated at C, the50 cm mark (Fig.)
According to the principle of moments
Moment of mass of coins about C’ = moment of mass of the rod about C’
10 g (45 – 12) = mg (50 – 45)

Question - 19 : - A solid sphere rolls down two different inclinedplanes of the same heights but different angles of inclination. Will it reachthe bottom with the same speed in each case ? Will it take longer to roll downone plane than the other ? If so, which one and why

Answer - 19 : -

Since sphere rolls down two inclined planes of sameheight, so velocity of sphere in both the cases is same.

Since θ is different in both the cases, so sphere will takelonger time in case of inclined plane having smaller inclination angle (0).

Question - 20 : - A hoop of radius 2 m weighs 100 kg. It rolls along ahorizontal floor so that its center of mass has a speed of 20 cm/s. How muchwork has to be done to stop it ?

Answer - 20 : - Here,  R = 2m, 100 kg
u = 20 cm/s = 0.2 m/s

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