A car wheel is rotating at 1200 revolutions p | vedclass

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(D) The initial angular velocity is $\omega_1 = 1200 \, rpm = \frac{1200}{60} \, rev/s = 20 \, rev/s$.The final angular velocity is $\omega_2 = 4500 \

Vedclass · Accepted Answer

$1980 \, deg/s^2$

Vedclass · Answer

(D) The initial angular velocity is $\omega_1 = 1200 \, rpm = \frac{1200}{60} \, rev/s = 20 \, rev/s$.The final angular velocity is $\omega_2 = 4500 \, rpm = \frac{4500}{60} \, rev/s = 75 \, rev/s$.The change in angular velocity is $\Delta \omega = \omega_2 - \omega_1 = 75 - 20 = 55 \, rev/s$.The time taken is $t = 10 \, s$.The angular acceleration in $rev/s^2$ is $\alpha = \frac{\Delta \omega}{t} = \frac{55}{10} = 5.5 \, rev/s^2$.To convert this to $deg/s^2$,we multiply by $360^\circ$ (since $1 \, rev = 360^\circ$):$\alpha = 5.5 	imes 360 \, deg/s^2 = 1980 \, deg/s^2$.

$A$ car wheel is rotating at $1200$ revolutions per minute. On pressing the accelerator,it reaches $4500$ revolutions per minute in $10 \, s$. The angular acceleration of the wheel is:

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