Posted by **lola** on Saturday, December 18, 2010 at 4:06pm.

A ball is placed at the top of a hill of incline 60.0 degrees. The ball is

allowed to roll for 12.0 seconds. How far will the ball roll during this

period? Neglect friction.

- phyics -
**bobpursley**, Saturday, December 18, 2010 at 4:23pm
Neglect friction, but the ball is rolling? What makes is roll? Remember, you said neglect friction.

Assuming there is friction, at least enough to rotate the ball, then the ball falls some vertical distance h.

This corresponds with incline distance L=h/sin60

1/2 m v^2+1/2 I w^2=mgh

but w=v/r, and I for a solid ball is 2/5 m r^2

1/2 m v^2+1/2 2/5 mv^2=mgh

v^2(1/2 +1/5)=gh

v= sqrt (gh/(3.5/5))

this is final velocity,so average velocity is 1/2 v or 1/2 sqrt (5gh/3.5)

how far did it go? in 12 seconds?

distance down the hill= v*t=6sqrt(5gh/3.5)

check my thinking.

- physics -
**drwls**, Saturday, December 18, 2010 at 5:16pm
If there is friction, and the ball rolls, the rate of acceleration depends upon the moment of inertia of the ball. Is it hollow or solid? These things male a difference.

It is quite unlikely that a ball will slip and fail to roll, but that seems to be what they want you to assume.

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