Posted by **Zoey** on Sunday, July 20, 2014 at 2:35pm.

A hoop of mass M = 4 kg and radius R = 0.4 m rolls without slipping down a hill, as shown in the figure. The lack of slipping means that when the center of mass of the hoop has speed v, the tangential speed of the hoop relative to the center of mass is also equal to vCM, since in that case the instantaneous speed is zero for the part of the hoop that is in contact with the ground (v − v = 0). Therefore, the angular speed of the rotating hoop is ω = vCM/R.

The initial speed of the hoop is vi = 2 m/s, and the hill has a height h = 3.7 m. What is the speed vf at the bottom of the hill?

(b) Replace the hoop with a bicycle wheel whose rim has mass M = 4 kg and radius R = 0.4 m, and whose hub has mass m = 1.3 kg, as shown in the figure. The spokes have negligible mass. What would the bicycle wheel's speed be at the bottom of the hill? (Assume that the wheel has the same initial speed and start at the same height as the hoop in part (a)).

## Answer this Question

## Related Questions

- science - A hoop of mass M = 4 kg and radius R = 0.4 m rolls without slipping ...
- Physics - A hoop of mass M = 3 kg and radius R = 0.5 m rolls without slipping ...
- Physics - 15. [1pt] A thin hoop of radius r = 0.59 m and mass M = 9.2 kg rolls ...
- Physics - What is the final velocity of a hoop that rolls without slipping down ...
- Physics - A 1-kg thin hoop with a 50-cm radius rolls down a 47° slope without ...
- Physics - A 1-kg thin hoop with a 50-cm radius rolls down a 47° slope without ...
- physics - As part of a kinetic sculpture, a 5.6 kg hoop with a radius of 3.8 m ...
- Physics Classical Mechanics - A hollow cylinder of outer radius R and mass M ...
- physics - A ball of mass .15 kg and a radius of .24 m is at the top of a 3.5 m ...
- Physics - A bowling ball of mass M and radius R is thrown along a level surface ...