Rotational Motion

  • #1
A 13 kg cylinder with a .54 m diameter rolls without slipping down a 30 degree incline from a height of 1.25 m.

a. If the cylinder has I = (mr^2)/2, what will its speed be at the incline's base?

After using the conservation of mechanical energy I got v = sqrt[(4/3)gh], which is 4.04 m/s.

b. What is its speed if it rolls from the same height down a 60 degree incline?

Now wouldn't the speed be the same as in part a since v is not dependent on the angle?

My major question is why are both speeds the same if the angle changes?

If my calculations are incorrect, please inform me. Thank you for any help.
 

Answers and Replies

  • #2
Doc Al
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Soaring Crane said:
After using the conservation of mechanical energy I got v = sqrt[(4/3)gh], which is 4.04 m/s.
Right!

b. What is its speed if it rolls from the same height down a 60 degree incline?

Now wouldn't the speed be the same as in part a since v is not dependent on the angle?
Right.

My major question is why are both speeds the same if the angle changes?
Since the initial PE is the same in both cases, the final speed will be the same. What does change is the acceleration down the incline. The smaller angle produces a smaller acceleration, but the distance is greater by the same factor, so the final speed remains the same.
 

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