The angular speed of a motor wheel is increas | vedclass

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(A) Given,initial angular frequency of the wheel,$f_0 = 1200 rpm = \frac{1200}{60} rps = 20 rps$.Initial angular velocity,$\omega_0 = 2 \pi f_0 = 2 \p

Vedclass · Accepted Answer

$4 \pi rad/s^2$

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(A) Given,initial angular frequency of the wheel,$f_0 = 1200 rpm = \frac{1200}{60} rps = 20 rps$.Initial angular velocity,$\omega_0 = 2 \pi f_0 = 2 \pi 	imes 20 = 40 \pi rad/s$.Final angular frequency,$f = 3120 rpm = \frac{3120}{60} rps = 52 rps$.Final angular velocity,$\omega = 2 \pi f = 2 \pi 	imes 52 = 104 \pi rad/s$.Time taken,$t = 16 s$.Using the equation of rotational motion,$\omega = \omega_0 + \alpha t$,where $\alpha$ is the angular acceleration.$\alpha = \frac{\omega - \omega_0}{t} = \frac{104 \pi - 40 \pi}{16} = \frac{64 \pi}{16} = 4 \pi rad/s^2$.

The angular speed of a motor wheel is increased from $1200 rpm$ to $3120 rpm$ in $16 s$. The angular acceleration of the motor wheel is

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