Question

A solid sphere of diameter 25 cm and mass 25kg rotates about an axis through its centre. Calculate its moment of inertia, if its angular velocity changes from 2 rad/s to 12 rad/s in 5 second. Also calculate the torque applied.
(Ans: I = 0.1562 kgm² τ = 0.3124 Nm)

Answer 1

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Here is Your Answer....!!!!
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⭐⭐Actually welcome to the concept of the ROTATIONAL DYNAMICS. .

⭐basically we know that ...

⭐moment of inertia(I cm) about its central axis of rotation of a Solid sphere is given as 2/5 Mr^2

⭐And torque is given as tau (T) = I alpha ...

⭐Where (alpha ) is the angular acceleration
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Hope it helps you...☺

Answer 2

Given,

radius  of  sphere,(R) = 25/2 cm  = 0.25/2 m

Mass(m) = 25 kg

We know that  moment  of  inertia  of a  sphere is  2/5 mR² where  m is  mass  of  sphere and  R  is its  radius .

Inertia  of  Sphere = 2/5 mR²

                               = 2/5 × 25 × (0.25/2)²

                                 = 10 × 0.25/2 × 0.25/2

                                 =  10/4 × 25/100 × 25/100

                                = 625/4000

                               = 0.15625 k-m²

Given that,its angular velocity changes from 2 rad/s to 12 rad/s in 5 second.

We  know that  rate of  change  in angular  velocity with respect  to time is  called angular  acceleration ∝.

∝ = (12- 2)/5 = 10/5 = 2 rad/s²

Again,

Torque = Inertia × ∝

             = 0.15625 × 2

             =  0.31250 N-m