A disc has mass m and radius R. How much tang | vedclass

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(A) The angular acceleration $\alpha$ is given by $\alpha = \frac{\omega}{t}$.The moment of inertia of a disc about its central axis is $I = \frac{1}{

Vedclass · Accepted Answer

$\frac{mR \omega}{2 t}$

Vedclass · Answer

(A) The angular acceleration $\alpha$ is given by $\alpha = \frac{\omega}{t}$.The moment of inertia of a disc about its central axis is $I = \frac{1}{2} mR^2$.The torque $	au$ required is $	au = I \alpha = (\frac{1}{2} mR^2) (\frac{\omega}{t}) = \frac{mR^2 \omega}{2t}$.Since torque is also given by $	au = F \cdot R$,where $F$ is the tangential force applied at the rim,$F \cdot R = \frac{mR^2 \omega}{2t}$.Solving for $F$,we get $F = \frac{mR \omega}{2t}$.

$A$ disc has mass $m$ and radius $R$. How much tangential force should be applied to the rim of the disc so as to rotate it with angular velocity $\omega$ in time $t$?

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