Problem # 1 (3+1+2+2+1-9 points) A thin-walled hollow sphere of mass M = 4.4 kg

Question

Problem # 1 (3+1+2+2+1-9 points) A thin-walled hollow sphere of mass M = 4.4 kg and radius R = 0.45 m rolls up the incline without slipping. The incline is at the angle = 350 with respect to horizontal. The sphere begins to roll up the incline with initial speed vo 2.5 m/s. The sphere moves up the incline and eventually stops. The moment of inertia of the hollow Vo 2 sphere is 1MR. Hint: use the mechanical energy conservation to answer (b) and Find: (a) the instantaneous angular velocity (in rad/sec) of the sphere right before it begins to roll up the incline (1 points); (b) how high above the ground (in m) the sphere rolls up the incline before it stops (2 points); (c) if the hollow sphere is replaced with a solid sphere of the same mass and radius, how high above the ground (in m) the solid sphere rolls up the incline before it stops (2 points). The moment of inertia of a solid sphere is 2 (d) explain why answers to (b) and (c) are different (1 point).

Explanation / Answer

(A) w = v / R = 2.5 / 0.45 = 5.6 rad/s

(B) KEi + PEi = KEf + PEf

(4.4 x 2.5^2)/2 + (2 x 4.4 x 0.45^2 / 3)(5.6^2)/2 = 0 + 4.4 x9.8 h

h = 0.53 m

(C) m v^2 /2 + I w^2 /2 = m gh

m v^2/2 + ( 2 m r^2/ 5)(v/r)^2 /2 = m gh

v^2 /2 + v^2 / 5 = g h

(2.5^2)(1/2 + 1/5) = 9.8 h

h =0.45 m

(d) speed is same for both so same translational Ke.

and rotational KE depends on moment of inertia.

moment of inertia is greater of hollow sphere so it had more KE.

that means it can get greater potential energy.

that means gerater height.

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