Answer to Question #249637 in Molecular Physics | Thermodynamics for Kelani

Question

A 2.30 kg hoop of radius 0.160 m is released from rest at point A in the figure below, its center of gravity a distance of 1.80 m above the ground. The hoop rolls without slipping to the bottom of an incline and back up to point B where it is launched vertically into the air. The hoop then rises to its maximum height h_max at point C.

(a)

At point B, find the hoop's translational speed v_B (in m/s).

___ m/s

(b)

At point B, find the hoop's rotational speed 𝜔_B (in rad/s).

___ rad/s

(c)

At point C, find the hoop's rotational speed 𝜔_C (in rad/s).

___ rad/s

(d)

At point C, find the maximum height h_max of the hoop's center of gravity (in m).

 ___ m

Answer 1

(a) Apply energy conservation:

$$\frac12 mv_B^2+\frac12I\omega^2=mgh,\\\space\\ \frac12 mv_B^2+\frac12(mr^2)\bigg(\frac{v_B}{r}\bigg)^2=mgh, \\\space\\ v_B^2m=mgh, \\\space\\ v_B=\sqrt{gh}=4.2\text{ m/s}.$$

(b) The rotational speed is

$$\omega_B=v_Br=0.672\text{ rad/s}.$$

(c) Assuming zero air resistance, the speed is the same, 0.672 rad/s.

(d) The maximum height is the same according to energy conservation, or 1.8 mm.