Cylinder rolling up inclined plane - Rolling without slipping

Oshada

1. The problem statement, all variables and given/known data



2. Relevant equations

All the usually relevant circular motion equations involving θ, I, ⍺, τ and ω

3. The attempt at a solution

I've tried to work from the energy of the cylinder (Using Ek = Ktrans + Krot) and equating the energy to its potential energy at the point it starts to roll down but got nowhere near the answer. I've done the force diagram, which had the weight force, normal force and friction. For c), I think I'm supposed to use τ = r x F and F is the friction force. But I'm not sure whether to just use linear acceleration formulae to obtain acm.

Any help is welcome!

ehild

It is all right that you tried energy conservation. But without seeing your work in detail, we cannot find the error in your solution.


ehild

Oshada

Sorry, here it is:

At the beginning: E = Ktrans + Krot (since Ep = 0) = 1/2 * m * v^2 + 1/2 * I * ω^2.
Using v = rω: E = (1/2*3.8*5.6^2) + (1/2*3.8*r^2*(5.6^2/r^2)
Therefore E = 120J (to two sig. figs)

When the cylinder starts to roll back; let x be the height at which the cylinder is motionless (therefore x = sin(θ)*4.2):
E = Pe (since Ek = 0) = 120J. (119.168)
Pe = mgh = 3.8*9.8*4.2*sin(θ) = 120 (119.168)

This gives me θ = 43.7 degrees. The answer is 34.8 degrees.

ehild

You miss a factor of 1/2 from the rotational energy. Erot=1/2 Iw^2, but I=1/2 Mr^2

ehild

Oshada

Oh my! That was so stupid on my part. Thank you very much! Also in part c), how does r x F work?

ehild

What do you mean? You know the acceleration, and ma=F(resultant). The forces acting on the cylinder along the slope are the component of gravity and the force of static friction. You need the expression for the static friction.

ehild

Oshada

So basically: ma = mgsin(θ) - Fr yes? But the answer is Fr = (M * acm)/2. I'm sure I'm missing something simple again...

ehild

So you can not use the angle of inclination. But there is an equation between torque and angular acceleration. Use that one.

ehild

Oshada

I⍺ = r x F is what I was going to use, but the sin(θ) still crops up!

ehild

No, why?

ehild

Oshada

From Fnet = mgsin(θ) - Fr?

ehild

It is really I⍺ = r x Fr , Fr is the force of friction. How is related the angular acceleration, alpha, to acm? r is the radius drawn to the bottom point of the cylinder, where it touches the slope. R is at right angle with Fr. You know everything.

ehild

Oshada

This might be a foolish question, but why isn't the weight component parallel to the plane included? I⍺ = r x Fr gives the right answer though :D

ehild

The force of gravity acts in the centre of mass. So its torque is zero with respect to the cm .

ehild

Oshada

Makes perfect sense, thank you very much!

ehild

You are welcome.

ehild