A solid metal ball starts from rest and rolls without slipping a distance of d =

Question

A solid metal ball starts from rest and rolls without slipping a distance of d = 4.4 m down a = 20° ramp. The ball has uniform density, a mass M = 4.9 kg and a radius R = 0.33 m.

1)

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

2)

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

J

3)

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

m/sec

4)

What is the magnitude of the frictional force on the sphere?

N

5)

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?

J

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Explanation / Answer

1) KE at any point = Translational KE + Rotational KE

= ( 0.5 . m . v^2 ) + (0.5 . I . w^2 )

I = (2/5) . m . r^2

w = v/r

so KE = 0.5 . m . v^2 + 0.5 . 0.4 . m . r^2 . v^2/r^2

KE = 0.7 .m . v^2

So the translational KE is 5/7 ths of the total KE.

2) At the bottom all the height energy lost has been converted into KE.

Of that KE 5/7 ths is translational KE.

Height energy lost = m . g. h = m .g . 4.4 . sin20

Translational KE = 5/7 . (4.9) . (9.8) . (4.4) .sin20 = 51.62 J

3) Just equate the 51.62 J with 0.5 . m . v^2 and solve for v. It comes to 4.59 m/s

4)

f = mgcos20 = (4.9) . (9.8) .cos20 = 45.12N

5) If there is no friction the ball will slide without rolling so there is no rotational KE to take into account.

So

Translational KE = mgh = (4.9) . (9.8) . (4.4) .sin20 = 72.26 J

[ m.g.h = 0.5 . m . v^2

v = ( 2 . 9.8 . 4.4 . sin 20) = 5.43 m/s]

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