A wheel of radius 30 cm is driven by a belt. | vedclass

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(B) The radius of the wheel is $r = 30 \ cm = 0.3 \ m$. The initial angular speed is $\omega_0 = 2 \ rev/s = 2 	imes 2\pi \ rad/s = 4\pi \ rad/s$. Th

Vedclass · Accepted Answer

$-0.95$

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(B) The radius of the wheel is $r = 30 \ cm = 0.3 \ m$. The initial angular speed is $\omega_0 = 2 \ rev/s = 2 	imes 2\pi \ rad/s = 4\pi \ rad/s$. The length of the belt passed is $s = 25 \ m$. The angular displacement $	heta$ is given by $	heta = \frac{s}{r} = \frac{25}{0.3} = 83.33 \ rad$. Since the wheel comes to rest,the final angular speed is $\omega = 0$. Using the kinematic equation $\omega^2 = \omega_0^2 + 2\alpha	heta$,we get: $0 = (4\pi)^2 + 2 	imes \alpha 	imes 83.33$. $\alpha = -\frac{16\pi^2}{166.66} \approx -\frac{157.91}{166.66} \approx -0.947 \ rad \ s^{-2}$. Rounding to two decimal places,the angular deceleration is $-0.95 \ rad \ s^{-2}$.

$A$ wheel of radius $30 \ cm$ is driven by a belt. Its initial angular speed is $2 \ rev/s$. Starting at this speed,it comes to rest after a length of $25 \ m$ of the belt has passed over the wheel. The angular deceleration of the wheel is ....... $rad \ s^{-2}$.

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