A marble rolls down an incline of 30 degrees. If the marble is initially at rest

Question

A marble rolls down an incline of 30 degrees. If the marble is initially at rest at the tip of the incline of length 2.0 meters, what will the speed of the marble be att the bottom of the incline if:

1) the incline is rough enough to prevent any slipping (it rolls)?

2) the incline is perfectly smooth (it slides without rolling)?

3) Why does the marble have different speeds when the total energy must be conserved in both cases? (clear argument is required).

Please show the equations used, thank you!

Explanation / Answer

Part 1)

The KE is partially rotational and partially translational

I for a sphere (the marble) is (2/5)mr^2 and angular speed = v/r

Since KE rotational = .5Iw^2 we can say that, for the marble...

KEr = (.5)(2/5)mr^2)(v^2/r^2) which simplifies to .2mv^2

So, by conservation of energy

mgh = .5mv^2 + .2mv^2 (mass cancels)

(9.8)(2)(sin 30) = (.7v^2)

v = 3.74 m/s

Part 2 (no rolling)

mgh = .5mv^2

(9.8)(2)(sin 30) = .5v^2

v = 4.43 m/s

Part 3)

The total energy is conserved. In the first case you have some of the KE in the translational movement of the ball and some in the rotational movement. In the second case, there is not rolling portion to take away some of the energy from the translational case. The translational case is what concerns the speed, and when you don't have rolling taking away some of the energy, the speed increases

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