Cylinder on inclined plane

[NAkid]

Homework Statement

A block of a certain material begins to slide on an inclined plane when the plane is inclined to an angle of 11.86o. If a solid cyclinder is fashioned from the same material, what will be the maximum angle at which it will roll without slipping on the plane (in degrees)?

Homework Equations

The Attempt at a Solution

drew a free body diagram, and have the following equations

mgsin(theta) + friction = ma
torque = friction*R = I(angular acceleration) --> friction=(I*angular acceleration)/R
mgsin(theta) + (I*angular acceleration)/R = ma
a=R*(angular acceleration) and I = .5MR^2
... substitute those in, but then I get

theta = (inverse sin)[(.5*a)/g] -- how do i solve for a?

 

[Shooting Star]

The Attempt at a Solution

drew a free body diagram, and have the following equations

mgsin(theta) + friction = ma
torque = friction*R = I(angular acceleration) --> friction=(I*angular acceleration)/R
mgsin(theta) + (I*angular acceleration)/R = ma
a=R*(angular acceleration) and I = .5MR^2
... substitute those in, but then I get

theta = (inverse sin)[(.5*a)/g] -- how do i solve for a?

Have you used the relation between $v$ and $\omega$?

 

[Moetasim]

I would approach this problem by first understanding the concept of rolling without slipping. In this case, the maximum angle at which the cylinder will roll without slipping is when the friction force is equal to the component of the weight of the cylinder down the inclined plane. This means that the cylinder will just start to slip and roll at the same time.

To solve for the maximum angle, we can use the equation for torque: torque = friction * radius. We know that the torque due to friction is equal to the weight of the cylinder times the radius of the cylinder. We can also use the equation for torque for a cylinder: torque = moment of inertia * angular acceleration.

Setting these two equations equal to each other, we get:

mgRsin(theta) = (1/2)MR^2 * alpha

Where alpha is the angular acceleration and theta is the angle of the inclined plane. We can rearrange this equation to solve for alpha:

alpha = (2mgRsin(theta))/MR^2

Now we can use the equation for acceleration for a rolling cylinder: a = R * alpha. Substituting in our value for alpha, we get:

a = (2mgRsin(theta))/MR

Finally, we can use this equation to solve for the maximum angle at which the cylinder will roll without slipping:

theta = sin^-1((aMR)/(2mg))

We can plug in the known values for mass, radius, and gravitational acceleration to solve for theta. The resulting value will be the maximum angle at which the cylinder will roll without slipping on the inclined plane.