Moment of inertia of a ring is $3 k g m^{2}$ . It is rotated for $20 s$ from its rest position by a torque of $6 N - m$ .Calculate the work done.

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Solution

$I = 3 k g m^{2}$ initial angular velocity = 0
t = 20 seconds
Now, Torque $= I α$
$6 = 3 \times α$
$∴ α = 2 r a d / s^{2}$

$Θ = w_{0} t + \frac{1}{2} α t^{2}$
$Θ = 0 \times t + \frac{1}{2} \times 2 \times 20^{2}$
$Θ = 400 r a d$

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Moment of inertia of a ring is 3 kg m2. It is rotated for 20 s from its rest position by a torque of 6 N−m.Calculate the work done.

I=3kgm2 initial angular velocity = 0 t = 20 seconds Now, Torque =Iα6=3×α∴α=2rad/s2Θ=w0t+12αt2Θ=0×t+12×2×202Θ=400rad

Moment of inertia of a ring is $3 k g m^{2}$ . It is rotated for $20 s$ from its rest position by a torque of $6 N - m$ .Calculate the work done.

$I = 3 k g m^{2}$ initial angular velocity = 0
t = 20 seconds
Now, Torque $= I α$
$6 = 3 \times α$
$∴ α = 2 r a d / s^{2}$

$Θ = w_{0} t + \frac{1}{2} α t^{2}$
$Θ = 0 \times t + \frac{1}{2} \times 2 \times 20^{2}$
$Θ = 400 r a d$