Moment of inertia of body about on axis is 4 kg m^2 . The body is initially at rest and a torque of 8 N m starts acting on it along the same axis . Work done by torque in 20 secs is : -

Moment of inertia of body about on axis is 4 kg m^2 .  The body is initially at rest and a torque of  8 N m  starts acting on it along the same axis . Work done by torque in 20 secs is . . . . . . . . . . : -

A ) 1600 J  ,                                                        B ) 3200 J  ,

C ) 4800 J ,                                                         D ) None of these ,

Correct Answer : -   B ) 3200 J  ,

Solution : -

Work – energy theorem  : -

∆ W.D by all forces = change in K.E  ,

∆ W.D  =  ∆ K.E  =  ( 1 / 2 ) I ( ωf )^2   -  ( 1 / 2 ) I ( ωi )^2    , . . . . . . . . . . .  eq. 1

∴    I = inertia  ,     ωf = final angular velocity  ,  ωi = initial angular velocity .

So , first we will find the value of ωf and ωi . 

ωi  =  0  ( rad / s ) ,  Because body is initially at rest .

ωf = ωi +  α t  . . . . . . . . . . . . . eq. 2

So ,  

Ʈ = I α  ,     ∴   Ʈ = torque  ,  α = angular acceleration .

 α = Ʈ / I  =  8 /4  =  2  ( rad / s^2 ) , 

and  time = t = 20 secs .

So put  in eq. 2  : -

ωf = 0 +  2 ( 20 )  =  40  ( rad / s )  .

So put value of  ωf  in eq. 1  : -

∆ W.D  =  ( 1 / 2 ) x 4 x ( 40 )^2    -   ( 1 / 2 ) x 4 x ( 0 )^2   ,

∆ W.D  =  ( 1 / 2 ) x 4 x ( 40 )^2    -  0  ,

∆ W.D  =  2 x ( 40 )^2    ,

∆ W.D  =  3200 J  ,  Answer  .

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