A solid sphere on top of a 30degree inclined plane has a radius R = 0.10 M and a

Question

A solid sphere on top of a 30degree inclined plane has a radius R = 0.10 M and a mass of 5 kg. The sphere is released from rest at a height h = 1 m. Use conservation of energy to find: The speed of the center of mass VCM right before it hits ground The angular momentum of the sphere before it hits ground The rotational energy of the sphere before it hits ground. The fraction of potential energy at the top of the inclined plane is converted into the rotational energy at the bottom of the inclined plane. (20 [points for (a), 10 point for (b), 5 points each for (c) and (d))

Explanation / Answer

Part A)

Conservation of Energy...

PE at the top will equal the sum of the translational KE and rotational KE

Rotational KE = .5Iw2

I for a solid sphere is 2/5(mr)2 and w = v/r, so w2 = v2/r2

Thus, KEr = .5(2/5)mr2v2/r2 which simplifies to .2mv2

Therefore, the energy formula is...

PE = KEt + KEr

mgh = .5mv2 + .2mv2 (mass cancels)

gh = .7v2

(9.8)(1) = .7v2

v = 3.74 m/s

Part B)

L = Iw

L = (2/5)(mr2)(v/r)

L = (2/5)(5)(.1)(3.74)

L = .748 kg m2/s

Part C)

KEr = .2mv2

KEr = (.2)(5)(3.74)2

KEr = 14.0 J

Part D)

KEr/PE

14/(5)(9.8)(1)

Fraction = .256

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