A hoop with a mass of 2.75 kg is rolling without slipping along a horizontal surface with a speed of 4.50 m/s when it starts down a ramp that makes an angle of 25.0° with the horizontal. What is the rotational kinetic energy of the hoop after it has rolled 3.00 m down the ramp?
A) 34.2 J
B) 22.4 J
C) 44.9 J
D) 62.0 J
E) This question cannot be answered without knowing the radius of the hoop.

Question

Quizplus · Accepted Answer

Firstly, we can use conservation of energy to find the hoop's speed when it reaches the bottom of the ramp:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
where $m$ is the mass of the hoop, $h$ is the height of the ramp, $v$ is the linear speed of the hoop at the bottom of the ramp, $I$ is the moment of inertia of the hoop, and $\omega$ is the angular speed of the hoop at the bottom of the ramp. Since the hoop is rolling without slipping, we know that $v = r\omega$, where $r$ is the radius of the hoop. Thus, we can rewrite the above equation as:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\frac{v^2}{r^2}$
Rearranging, we get:
$v^2 = \frac{2mgh}{m + \frac{I}{r^2}}$
We can find the moment of inertia of the hoop about its center of mass using the formula for the moment of inertia of a thin, uniform hoop:
$I = mr^2$
Plugging in the given values, we get:
$v^2 = \frac{2(2.75	ext{ kg})(9.81	ext{ m/s}^2)(3.00	ext{ m})}{2.75	ext{ kg} + \frac{(2.75	ext{ kg})(1.0^2)}{2}}$
$v = 6.587	ext{ m/s}$
The rotational kinetic energy of the hoop can then be found using the formula:
$K_{rot} = \frac{1}{2}I\omega^2$
Since the hoop is rolling without slipping, we know that $\omega = \frac{v}{r}$. Plugging in the given values, we get:
$K_{rot} = \frac{1}{2}(2.75	ext{ kg})(1.0^2)\left(\frac{6.587	ext{ m/s}}{1.0	ext{ m}}ight)^2$
$K_{rot} = 44.94	ext{ J} \approx 44.9	ext{ J}$

Therefore, the best choice is (C).

A Hoop with a Mass of 2