Question

A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.

Answer 1

Answer:

A) Must be done 19806.62 joules of work.

B) The average power is 1320.44 Watts.

Explanation:

A) First, we're going to use the work-energy theorem that states total work ($W$) done on an object is equal to the change in its kinetic energy ($\Delta K$):

$W=\Delta K = K_{f}-K_{i}$ (1)

So, all we must do is to find the change on kinetic energy. Because we're working with rotational body, we should use the equation $K=\frac{I\omega^{2}}{2}$ for the kinetic energy so:

$\Delta K=\frac{I(\omega_{f})^{2}}{2}-\frac{I(\omega_{i})^{2}}{2}$ (2)

with $\omega_{i}$ the initial angular velocity, $\omega_{f}$ the final angular velocity (is zero because the wheel stops) and I the moment of inertia that for a thin hoop is $I=MR^{2}$, using those on (2)

$\Delta K=0-\frac{MR^{2}(\omega_{i})^{2}}{2}$ (3)

By (3) on (1):

$W= \frac{MR^{2}(\omega_{i})^{2}}{2} = \frac{(32.0)(1.2)^{2}(29.32)^{2}}{2}$

$W=19806.62\,J$

B) Average power is work done divided by the time interval:

$P=\frac{W}{\Delta t}=\frac{19806.62}{15.0}$

$P=1320.44\,W$

NOTE: We use the relation $1rpm*\frac{2\pi}{60s}=\frac{rad}{s}$ to convert 280 rev/min(rpm) to 29.32 rad/s