Sphere on Incline A solid sphere of uniform density starts from rest and rolls w

Question

Sphere on Incline

A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 3.7 m down a = 38° incline. The sphere has a mass M = 3.8 kg and a radius R = 0.28 m.

1) Of the total kinetic energy of the sphere, what fraction is translational?

KE tran/KEtotal =

2)What is the translational kinetic energy of the sphere when it reaches the bottom of the incline?

KE tran =    J

3)What is the translational speed of the sphere as it reaches the bottom of the ramp?

v =     m/s

4)Now let's change the problem a little.

Suppose now that there is no frictional force between the sphere and the incline. Now, what is the translational kinetic energy of the sphere at the bottom of the incline?

KE tran =    J

Explanation / Answer

(1) Applying energy conservation,

initial PE + KE = final PE + KE

m g h + 0 = 0 + (m v^2 /2 + I w^2 / 2)

m g d sin38 = m v^2 / 2 + (2 m r^2 / 5) ( v/r)^2 / 2

g d sin38 = v^2 / 2 + v^2 / 5 = 7 v^2 / 10

v^2 = 10 x 9.8 x 3.7 x sin38 / 7 = 31.89

v = 5.65 m/s

total KE = 7 m v^2 / 10 = 7 x 3.8 x 5.65^2 / 10 = 84.83 J

translational KE = m v^2 /2 = 3.8 x 5.65^2 / 2 = 60.65 J

fration = 60.65 / 84.83 = 0.715

(2) trans KE = 60.65 J

(3) v = 5.65 m/s

(4) when there is no friction then

rotational Ke = 0

trans KE = total energy = m g d sin38 = 84.83 J

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