The angular speed of a truck wheel is increas | vedclass

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(B) Given:Initial angular speed,$\omega_1 = 900 \, rpm = \frac{900}{60} \, rev/s = 15 \, rev/s$.Final angular speed,$\omega_2 = 2460 \, rpm = \frac{24

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(B) Given:Initial angular speed,$\omega_1 = 900 \, rpm = \frac{900}{60} \, rev/s = 15 \, rev/s$.Final angular speed,$\omega_2 = 2460 \, rpm = \frac{2460}{60} \, rev/s = 41 \, rev/s$.Time interval,$t = 26 \, s$.Since the angular acceleration is uniform,the average angular speed is $\omega_{avg} = \frac{\omega_1 + \omega_2}{2}$.$\omega_{avg} = \frac{15 + 41}{2} = \frac{56}{2} = 28 \, rev/s$.The total number of revolutions $N$ is given by the product of average angular speed and time:$N = \omega_{avg} 	imes t = 28 \, rev/s 	imes 26 \, s = 728 \, revolutions$.

The angular speed of a truck wheel is increased from $900 \, rpm$ to $2460 \, rpm$ in $26 \, s$. The number of revolutions made by the truck engine during this time is (assuming the acceleration to be uniform).

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