Rotational Motion (thin hollow pipe)

[Bones]

Homework Statement

thin, hollow 60.0 g section of pipe of radius 14.0 cm starts rolling (from rest) down a 17.5° incline 5.60 m long.
(a) If the pipe rolls without slipping, what will be its speed at the base of the incline?
(b) What will be its total kinetic energy at the base of the incline?
(c) What minimum value must the coefficient of static friction have if the pipe is not to slip?

Homework Equations

The Attempt at a Solution

I got a) V=sqrt(gh) V=4.06m/s and b) KE=mv^2=mgh KE=0.989J which are correct.
c) I need help with this one, I am not sure where to begin.

 

[Redbelly98]

This is a tricky question.

Without thinking this through completely, I'll suggest drawing a force diagram for the piple, including friction. Then some equations involving Newton's 2nd Law, and the torque-rotation version of Newton's 2nd Law.

 

[thomate1]

For part c), we can use the equation for the minimum coefficient of static friction (μs) to prevent slipping, which is:

μs = tan(θ)

Where θ is the angle of the incline, in this case 17.5°.

Therefore, μs = tan(17.5°) = 0.315.

This means that the coefficient of static friction between the pipe and the incline must be at least 0.315 in order for the pipe to not slip down the incline.