# Rolling Down an Off Center Cylinder on an Incline

• Silviu
In summary, the cylinder starts to roll down at the angle ##\alpha=\beta/2## where the slope of the incline is maximal.

## Homework Statement

A cylinder of radius a and mass m contains a point mass, also of mass m, located a distance ##a/2## from the symmetry axis. The cylinder is placed on an incline, which is initially horizontal, but is very slowly raised. Assuming the cylinder cannot slide on the incline, at what inclination angle ##\alpha## does the cylinder begin to roll down?

## The Attempt at a Solution

I tried to write the torque around the contact point, with ##\theta## being the angle where the point mass is located, from the line connecting the contact point with the center of the cylinder. I have this: $$\tau = mg(a-a/2 cos(\theta))sin(\alpha)+mga sin(\alpha)=mga(2- cos(\theta)/2)sin(\alpha)$$ But i don't know what condition to put. I know I have to use the no slipping, which would be $$x=\theta a$$ where x is the distance along the plane, but I am not sure what equation involving x should I use.

Silviu said:

## Homework Statement

A cylinder of radius a and mass m contains a point mass, also of mass m, located a distance ##a/2## from the symmetry axis. The cylinder is placed on an incline, which is initially horizontal, but is very slowly raised. Assuming the cylinder cannot slide on the incline, at what inclination angle ##\alpha## does the cylinder begin to roll down?

## The Attempt at a Solution

I tried to write the torque around the contact point, with ##\theta## being the angle where the point mass is located, from the line connecting the contact point with the center of the cylinder. I have this: $$\tau = mg(a-a/2 cos(\theta))sin(\alpha)+mga sin(\alpha)=mga(2- cos(\theta)/2)sin(\alpha)$$ But i don't know what condition to put. I know I have to use the no slipping, which would be $$x=\theta a$$ where x is the distance along the plane, but I am not sure what equation involving x should I use.
I think that if you draw the correct diagram and write the related equation, the problem will almost solve itself. First draw the cylinder on an inclined plane. Show the point mass at distance ##a/2## from the axis of symmetry. Show the radius connecting the axis of symmetry and point mass, and the angle ##\beta## between that line and the vertical. Write the equation for the net torque about the point of contact in terms of ##\alpha## and ##\beta##. Now, in terms of angle ##\beta##, what is the condition which causes the cylinder to start rolling? Please show us your diagram and your equations.

Silviu said:
$$\tau = mg(a-a/2 cos(\theta))sin(\alpha)+mga sin(\alpha)=mga(2- cos(\theta)/2)sin(\alpha)$$
Since you want a net torque of zero, you can easily turn that into a relationship between the two angles and find the value of θ which maximises α. However, I do not think your equation is correct. Please explain how you arrive at it.

It might be easier to think in terms of the common mass centre. At equilibrium, where must that be in relation to the contact point?
As the slope increases, what happens to the horizontal displacement from the cylinder's centre to the common mass centre?