A \(31.4-\mathrm{kg}\) wheel with radius \(1.21 \mathrm{~m}\) is rotating at 283 rev/min. It must be brought to a stop in \(14.8 \mathrm{~s}\). Find the required average power. Assume the wheel to be a thin hoop.

Short Answer

Expert verified
The required average power to stop the wheel is 1125 Watts.

Step by step solution

01

Calculate the initial angular velocity

Firstly, we have to convert the rotational speed from rev/min to rad/sec. 1 rev = \(2π\) radian and 1 minute = 60 seconds, so the initial angular velocity ω = \(283 rev/min \times \frac{2π rad}{1 rev} \times \frac{1 min}{60 sec} = 29.6 rad/sec\)
02

Calculate the Moment of Inertia

The moment of inertia I for a hoop of mass m and radius r is \(I = m * r^2\). So, \(I = 31.4 kg * (1.1 m)^2 = 38.4 kg.m^2).
03

Calculate the initial Rotational Kinetic Energy

A body with moment of inertia I and angular velocity ω has the rotational kinetic energy KE = \(0.5 * I * ω^2\). So the initial rotational kinetic energy KE = \(0.5 * 38.4 kg.m^2 * (29.6 rad/sec)^2 = 16625 J\)
04

Calculate the required Power

The power required P to do work W in a time t is \(P = \frac{W}{t}\). In stopping the wheel, we do work equal to its initial rotational kinetic energy. So the required power P = \( \frac{16625 J}{14.8 s} = 1125 W\).

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