# Ball down hill w/ rotational motion

1. Nov 19, 2011

### QuarkCharmer

1. The problem statement, all variables and given/known data
A solid ball is released from rest and slides down a hillside that slopes downward at an angle 69.0 degrees from the horizontal.

What minimum value must the coefficient of static friction between the hill and ball surfaces have for no slipping to occur?

2. Relevant equations

3. The attempt at a solution
I'm not really sure what they mean here. If the ball is to roll down the hill, then it cannot "slide" at all, otherwise it would not roll. They don't give me any numbers to solve this with, and it's not a symbolic answer because the problem does not say "give your solution in terms of m,g,θ, et al.

What do they want me to do for this problem?

What I did was treat the ball as a box, and find the $μ_{s}$ like so:

I said that parallel to, and down the hill was the increasing x axis. Then I summed up the forces, applied F=ma, so solve:

$$mgsin(69)-f_{s}=ma$$
$$f_{s} = μ_{s}N = μ_{s}(mgcos(69))$$
$$mgsin(69)-μ_{s}(mgcos(69)=ma$$
Because it's not slipping, acceleration is zero, thus ma = 0
$$mgsin(69)-μ_{s}(mgcos(69)=0$$
$$mgsin(69) =μ_{s}(mgcos(69)$$
$$μ_{s} = \frac{mgsin(69)}{mgcos(69)}$$
$$μ_{s} = tan(69)$$

Does that make sense?

Last edited: Nov 19, 2011
2. Nov 19, 2011

### QuarkCharmer

Apparently this is incorrect. How else can I tackle this problem? I don't understand why this won't work just as well as summing up the torque and setting them to zero.

Thanks.

Last edited: Nov 19, 2011
3. Nov 19, 2011

### Staff: Mentor

You need to analyze both the rotational and the translational motion of the sphere. The friction affects both.

(Despite the sloppy wording of the problem statement, the sphere does not 'slide' down the hill. It rolls.)