Cylinder rolling up inclined plane - Rolling without slipping

[Oshada]

Homework Statement

Homework Equations

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

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]

Using v = rω: E = (1/2*3.8*5.6^2) + (1/2*3.8*r^2*(5.6^2/r^2)

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]

Also in part c), how does r x F work?

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]

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

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