I couldn\'t figure out part b. Please help. thank you! A hoop of mass M 2 kg and

Question

I couldn't figure out part b. Please help. thank you!

A hoop of mass M 2 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 ( 0). Therefore, the angular speed of the rotating hoop is -van/R of rim relative to center of mass o of center of mass (a) The initial speed of the hoop is - 3 m/s, and the hill has a height h-4.2 rm. What is the speed vr at the bottom of the hill? Vf 7.08 (b) Replace the hoop with a bicycle wheel whose rim has mass M - 2 kg and radius R 0.4 m, and whose hub has mass m - 1.5 kg, as shown in the figure. The spokes have negligible mass. What would the IT/s 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)) Vr- x m/s Additional Materials

Explanation / Answer

The only difference in the second part is in the total mass of the wheel which would be (M+m); the moment of inertia would however remain the same ie. MR2 as the extra mass m is concentrated near the axis of rotation and so doesn't contribute to the torque.

Therefore, in balancing the energy terms, the kinetic energy terms would remain the same (ie. 1/2 Iw2 where I is the moment of inertia and w is the angular velocity) and the gravitational potential energy term would become (M+m)gh.

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