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Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

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, I1 = mr

The moment of inertia of the solid sphere about an axis passing through its centre, I2 = mr

We have the relation, 

τ = Iα 

where, 

α = Angular acceleration, 

τ = Torque, 

I = Moment of inertia, 

For the hollow cylinder, τ1 = Iα1

For the solid sphere, τn = Iαn

As an equal torque is applied to both the bodies, ττ2, 

∴        =   =  

            α2 > α1                                      ...(i) 

Now, using the relation, 

ω = ω0 + α

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)

ω2 > ω

Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.
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To maintain a rotor at a uniform angular speed of 200 rad s–1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine?
(Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100 % efficient. 

Given,

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 × 10


  = 36 kW 

Hence, the power required by the engine is 36 kW. 
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(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.

(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
 

(a) 

Initial angular velocity, ω1= 40 rev/min 

Final angular velocity = ω

The moment of inertia of the boy with stretched hands = I

The moment of inertia of the boy with folded hands = I

The two moments of inertia are related as, 

                         I2 =  I1
Since no external force acts on the boy, the angular momentum L is a constant. 

Hence, for the two situations, 

            I2ω2  =  I1 ω

                ω2 = ω

                      


(b) 

Final K.E. = 2.5 Initial K.E.

Final kinetic rotation, EF = (1/2) I2 ω2

Initial kinetic rotation, EI =  (1/2) I1 ω12 

     = I2 ω2/  I1 ω1

         = I1 (100)2 / I1 (40)

         = 2.5 

Therefore, 

EF = 2.5 E

The increase in the rotational kinetic energy is attributed to the internal energy of the boy. 
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A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.


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 = mr

  = 3 × (0.4)2 

  = 0.48 kg m


Torque, τ = F × r 

             =  30 × 0.4 

             = 12 Nm 

For angular acceleration α, torque is also given by the relation, 

   τ = Iα 

   α = τ / I 

     =  12 / 0.48

     = 25 rad s-2 

Linear acceleration = τα 

                         = 0.4 × 25

                         = 10 m s–2 

This is the required linear acceleration of the rope. 
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A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s–1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

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, 

   I = mr2 / 2 

     = (1/2) × 20 × (0.25)

     = 0.625 kg m

∴ Kinetic energy = (1/2) I ω

                      = (1/2) × 6.25 × (100)2 

                      = 3125 J 

Therefore,

Angular momentum, 
L = Iω 

                               = 6.25 × 100 

                               = 62.5 Js 

That is, there is a magnitude of 62.5 Js angular momentum of the cylinder about it's axis. 
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