Cylinder rolling down the hill

In summary: You would then use the usual moment of inertia for rolling about the mass centre. That would be the same in each of your two cases.You are correct in taking the axis of rotation to be the point of contact. That is where the cylinder is not slipping. In the first case it is just more work to use that approach because the torque due to gravity at the mass centre is zero, so you would have to add the torque due to gravity at the end at a distance r from the centre.I used in both cases the parallel axis theorem.Yes, but in the second case you applied it to a body about its mass centre, so the parallel axis term is zero.This is not the same as your original post
  • #1
RiotRick
42
0

Homework Statement


rotationProblem.JPG

A solid cylinder rolls down an inclined plane because of stiction. How does angular velocity change over time?
Given:
##m,R,g,\alpha,\mu## where ##\mu## is the friction coefficient

Homework Equations


[/B]
##J_{cm} = 1/2 m*r^2## moment of inertia
m = mass of the cylinder
Parallel Axis theorem: ##J=J_{cm}+mr^2##
## M = I*d\omega/dt ## = Torque

The Attempt at a Solution


First I thought I take the torque at the center of mass pulling it down
##R*mg*sin(\alpha)=d\omega/dt*(J_{cm}+mr^2)##
which gives me as an end result:
## d\omega/dt = \frac{2*g*sin(\alpha)}{3*r} ##
= ## \omega = \frac{2*g*sin(\alpha)}{3*r}*t ##
Then I thought it would have been simpler if I looked at the point of friction:
## M/J = r \times F / J = r*F_{friction}/J = \mu*r*mg*cos(\alpha)/J=d\omega/dt ##
=## \frac{2*\mu*g*cos(\alpha)}{3r}=d\omega/dt ##
=## \frac{2*\mu*g*cos(\alpha)}{3r}*t=\omega ##

Are both correct, one of them or both garbage?
 

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  • #2
You cannot just ignore the change in linear velocity.

Edit: Also, is the cylinder assumed to be rolling without slipping or do you need to take slipping into account?
 
  • #3
RiotRick said:
First I thought I take the torque at the center of mass pulling it down
It does not seem to me that is what you did. You appear to have taken moments about the point of contact. In particular, you had to use the parallel axis theorem to get the MoI.
This is a good approach because it avoids dealing with the frictional force.
RiotRick said:
Then I thought it would have been simpler if I looked at the point of friction:
Which is what I think you did the first time. If you mean to consider the torque exerted by friction then you would now be taking moments about the mass centre, so the MoI is just ½mr2, but somehow you still got a 3 in there.
You will need another equation relating the frictional force to the linear acceleration, and another relating rotational to linear acceleration when rolling.
Also, you seem to have applied this popular but erroneous equation:
Static frictional force = coefficient of static friction x normal force.
That contains a serious flaw. Can you spot it?
Orodruin said:
is the cylinder assumed to be rolling without slipping or do you need to take slipping into account?
It says "stiction", which I take to mean static friction.
 
  • #4
Thank you both for the quick answer. It's rolling without slipping

@haruspex: So you think the first answer is correct?
I used in both cases the parallel axis theorem. So you mean in the 2nd case it's just the moment of inertia? Why is that? Because the rotation axis never falls together with the "ground"?
haruspex said:
That contains a serious flaw. Can you spot it?
No. I've never heard that there is something fishy with it
 
  • #5
RiotRick said:
No. I've never heard that there is something fishy with it
The coefficient of static friction tells you how hard you can push something before it "breaks loose" and starts slipping.

Say, for instance, you have a 50 kg box on a floor with a coefficient of static friction of 0.1 and gravity of 10 meters/sec^2. You rapidly calculate that you need a horizontal force of 50 * 0.1 * 10 = 50 N to overcome static friction and get it moving.

What is the force of static friction if you push the box horizontally with only 40 N?
 
  • #6
RiotRick said:
in the 2nd case it's just the moment of inertia?
If you are taking moments about the mass centre then the torque you calculate is producing an angular acceleration about the mass centre. So no parallel axis involved.
 

Related to Cylinder rolling down the hill

1. What is the force that causes a cylinder to roll down a hill?

The force that causes a cylinder to roll down a hill is gravity. As the cylinder gains height on the hill, it also gains potential energy. When released, this potential energy is converted into kinetic energy, causing the cylinder to roll down the hill.

2. How does the shape of the cylinder affect its rolling motion down the hill?

The shape of the cylinder can affect its rolling motion down the hill in a few ways. A wider cylinder will have a larger surface area in contact with the ground, creating more friction and slowing its descent. A taller cylinder may have a higher center of mass, making it more prone to tipping over. A perfectly round cylinder will have a consistent rolling motion, while a cylinder with uneven sides may experience a wobbly or erratic motion.

3. What role does friction play in the rolling motion of a cylinder down a hill?

Friction plays a significant role in the rolling motion of a cylinder down a hill. As the cylinder rolls, its bottom surface comes into contact with the ground. This contact creates friction, which can either aid or hinder the cylinder's motion. If the ground is rough, the friction will slow the cylinder down. If the ground is smooth, the cylinder will roll faster due to less friction.

4. How does the mass of the cylinder affect its rolling speed down the hill?

The mass of the cylinder can affect its rolling speed down the hill in two ways. First, a heavier cylinder will have more potential energy when at the top of the hill, resulting in a faster roll down. Second, a heavier cylinder will have more inertia, making it more difficult to change its direction or speed. This means that a heavier cylinder may roll down the hill more steadily and consistently, while a lighter cylinder may be more easily influenced by external forces.

5. Can a cylinder roll uphill?

In most cases, a cylinder cannot roll uphill. This is because the force of gravity is always pulling objects towards the center of the Earth, making it difficult for an object to move against this force. However, in certain situations, such as a cylinder rolling up a ramp, the force of gravity can be overcome by other forces, such as the force of the ramp pushing the cylinder upwards. In these cases, a cylinder can roll uphill, but it requires a significant amount of external force.

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