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(C) Given: Initial angular velocity $\omega_0 = 0$,Angular acceleration $\alpha = 8 \ rad \ s^{-2}$.$1$. For $t = 5 \ s$: The angular displacement is

Vedclass · Accepted Answer

$\frac{100}{2\pi }, \frac{48}{2\pi }, \frac{56}{2\pi }$

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(C) Given: Initial angular velocity $\omega_0 = 0$,Angular acceleration $\alpha = 8 \ rad \ s^{-2}$.$1$. For $t = 5 \ s$: The angular displacement is $	heta_5 = \omega_0 t + \frac{1}{2} \alpha t^2 = 0(5) + \frac{1}{2}(8)(5)^2 = 100 \ rad$.The number of revolutions in $5 \ s$ is $n_5 = \frac{	heta_5}{2\pi} = \frac{100}{2\pi}$.$2$. For $t = 6 \ s$: The angular displacement is $	heta_6 = \omega_0 t + \frac{1}{2} \alpha t^2 = 0(6) + \frac{1}{2}(8)(6)^2 = 144 \ rad$.The number of revolutions in the $6^{th}$ second is $\Delta n = \frac{	heta_6 - 	heta_5}{2\pi} = \frac{144 - 100}{2\pi} = \frac{44}{2\pi}$.$3$. After $6 \ s$,torque is zero,so $\alpha = 0$. The angular velocity at $t = 6 \ s$ is $\omega_6 = \omega_0 + \alpha t = 0 + (8)(6) = 48 \ rad \ s^{-1}$.In the $7^{th}$ second (i.e.,from $t=6$ to $t=7$),the angular displacement is $	heta_7 = \omega_6 	imes \Delta t = 48 	imes 1 = 48 \ rad$.The number of revolutions in the $7^{th}$ second is $\frac{48}{2\pi}$.

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