A hoop is rolling without slipping along a horizontal surface with a speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal. What is the speed of the hoop after it has rolled 3.00 m up the ramp?
A) 4.22 m/s
B) 1.91 m/s
C) 2.06 m/s
D) 3.79 m/s
E) 8.02 m/s

Question

Quizplus · Accepted Answer

First, we need to find the initial energy of the hoop. Since it is rolling without slipping, we can use the formula $E=\frac{1}{2}(I_{cm}+mR^2)\omega^2+\frac{1}{2}mv^2$, where $I_{cm}$ is the moment of inertia of the hoop about its center of mass, $m$ is the mass of the hoop, $R$ is the radius of the hoop, $\omega$ is the angular velocity of the hoop (which is related to the linear speed $v$ by $\omega=\frac{v}{R}$), and $v$ is the speed of the hoop.

For a uniform hoop, $I_{cm}=\frac{1}{2}mR^2$, so the formula simplifies to $E=\frac{3}{4}mv^2$. Plugging in $v=5.50	ext{ m/s}$, we find that the initial energy is $E_i=\frac{3}{4}m(5.50	ext{ m/s})^2=\frac{75}{4} m	ext{ m}^2/	ext{s}^2$.

As the hoop rolls up the ramp, its potential energy increases and its kinetic energy decreases due to the work done by gravity. When it reaches the top of the ramp, all of its initial kinetic energy will have been converted into potential energy, so we can set the final kinetic energy equal to zero and solve for the final speed:

\begin{align*}
E_i&=E_f\
\frac{3}{4}mv_i^2&=mg(h+\frac{1}{2}R\sin	heta)\
v_f&=\sqrt{\frac{4}{3}g(h+\frac{1}{2}R\sin	heta)}\
&=\sqrt{\frac{4}{3}\cdot9.81	ext{ m/s}^2\cdot(3.00	ext{ m}+\frac{1}{2}(0.20	ext{ m})\sin(25.0^\circ))}\
&=\boxed{	extbf{(A) }4.22	ext{ m/s}}
\end{align*}

A Hoop Is Rolling Without Slipping Along a Horizontal Surface