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Question

Retarding torque is constant. Therefore, angular retardation say a will also be constant. Applying

$\omega^{2}=\omega_{0}^{2}-2 \alpha 	heta \quad

Vedclass · Accepted Answer

$\frac{n}{3}$

Vedclass · Answer

Retarding torque is constant. Therefore, angular retardation say a will also be constant. Applying

$\omega^{2}=\omega_{0}^{2}-2 \alpha 	heta \quad$ we get

$\left[\frac{\omega_{0}}{2}ight]^{2}=\omega_{0}^{2}-2 \alpha 	heta_{1}$          $...(i)$

and $\quad 0=\left[\frac{\omega_{0}}{2}ight]^{2}-2 \alpha 	heta_{2}$          $...(ii)$

Solving equations $(i)$ and $(ii)$ we get $	heta_{2}=\frac{	heta_{1}}{3}$

Therefore, the disc will make $\frac{\mathrm{n}}{3}$ more rotations before coming to rest.

A disc is rotating with an angular velocity $\omega_0$. A constant retarding torque is applied on it to stop the disc. The angular velocity becomes $\frac{{{\omega _0}}}{2}$ after $n$ rotations. How many more rotations will it make before coming to rest

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