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

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(B) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} MR^2$.Using the equation of rotational motion,$\omega = \omega_0 + \alp

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

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(B) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} MR^2$.Using the equation of rotational motion,$\omega = \omega_0 + \alpha t$. Since the disc starts from rest,$\omega_0 = 0$,so $\alpha = \frac{\omega}{t}$.The torque $	au$ required is given by $	au = I \alpha = (\frac{1}{2} MR^2) 	imes (\frac{\omega}{t}) = \frac{MR^2 \omega}{2t}$.Since the force $F$ is applied tangentially at the rim,the torque is $	au = F 	imes R$.Equating the two expressions for torque: $F 	imes R = \frac{MR^2 \omega}{2t}$.Therefore,the tangential force is $F = \frac{MR \omega}{2t}$.

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

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