A hoop (I = MR²)of mass 2.0 kg and radius 0.50 m is rolling at a center-of-mass speed of 15 m/s.An external force does 750 J of work on the hoop.What is the new speed of the center of mass of the hoop? A)19 m/s B)22 m/s C)24 m/s D)27 m/s E)68 m/s

Question

Quizplus · Accepted Answer

The initial kinetic energy of the hoop is
$K_i = \frac{1}{2} I \omega^2 = \frac{1}{2} (2/3) mR^2 \omega^2 = \frac{1}{2} (2/3) (2.0 	ext{ kg})(0.50 	ext{ m})^2 (\frac{v_{	ext{cm}}}{R})^2 = \frac{1}{3} (2.0 	ext{ kg})(15 	ext{ m/s})^2 = 675 	ext{ J}$.
After the external force does 750 J of work on the hoop, the total kinetic energy of the hoop is $K_f = K_i + 750 	ext{ J}$.
Since no non-conservative forces do work on the hoop, the total mechanical energy of the system is conserved: $K_f + U_f = K_i + U_i$, where $U_i = mgh_i = 0$ (we can choose any height as the reference level for gravitational potential energy). Therefore, $K_f = K_i + 750 	ext{ J}$.
Solving for $v_{	ext{cm}}$ in $K_f = \frac{1}{2} I (\frac{v_{	ext{cm}}}{R})^2$ gives $v_{	ext{cm}} = \sqrt{\frac{2K_f}{I}} = \sqrt{\frac{4(K_i +750 	ext{ J})}{I}} = \sqrt{\frac{4(675 	ext{ J}+750 	ext{ J})}{(2/3) mR^2}} \approx 24 	ext{ m/s}$. Therefore, the answer is (C).

A Hoop (I = MR2)of Mass 2

A Hoop (I = MR²)of Mass 2