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(A) From the kinematic equations of circular motion, we have $\omega = \omega_0 + \alpha t$. Given $\omega_0 = 0$ at $t=0$, at $t=2 \ s$, $\omega = 5

Vedclass · Accepted Answer

$1000/\pi$

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(A) From the kinematic equations of circular motion, we have $\omega = \omega_0 + \alpha t$. Given $\omega_0 = 0$ at $t=0$, at $t=2 \ s$, $\omega = 5 \ rad/s$. $5 = 0 + \alpha 	imes 2 \Rightarrow \alpha = 2.5 \ rad/s^2$. For the interval $t=0 \ s$ to $t=20 \ s$, the angular displacement $	heta_1$ is: $	heta_1 = \omega_0 t + \frac{1}{2} \alpha t^2 = 0 + \frac{1}{2} 	imes 2.5 	imes (20)^2 = 500 \ rad$. The angular velocity at $t=20 \ s$ is $\omega_{20} = \omega_0 + \alpha 	imes 20 = 0 + 2.5 	imes 20 = 50 \ rad/s$. For the interval $t=20 \ s$ to $t=50 \ s$, the acceleration is zero, so the angular velocity remains constant at $50 \ rad/s$. The angular displacement $	heta_2$ is: $	heta_2 = \omega_{20} 	imes \Delta t = 50 	imes (50 - 20) = 50 	imes 30 = 1500 \ rad$. Total angular displacement $	heta = 	heta_1 + 	heta_2 = 500 + 1500 = 2000 \ rad$. Number of revolutions $n = \frac{	heta}{2\pi} = \frac{2000}{2\pi} = \frac{1000}{\pi}$.

$A$ wheel undergoes a constant angular acceleration from time $t=0$ to $t=20 \ s$ and thereafter angular acceleration is zero. If angular velocity at $t=2 \ s$ is found to be $5 \ rad/s$, then the number of revolutions made by the wheel in the time interval $t=0 \ s$ to $t=50 \ s$ is

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