What Angle Prevents a Cylinder from Slipping on an Inclined Plane?

AI Thread Summary
A block begins to slide on an inclined plane at an angle of 11.86 degrees, prompting the question of the maximum angle for a solid cylinder made from the same material to roll without slipping. The discussion involves applying equations of motion, including forces and torque, to derive the relationship between the angle of inclination, acceleration, and friction. A free body diagram is utilized to establish the equations mgsin(theta) + friction = ma and torque = friction*R = I(angular acceleration). The challenge arises in solving for acceleration (a) to find the angle theta using the equation theta = (inverse sin)[(.5*a)/g]. The conversation also hints at the importance of relating linear velocity and angular velocity in the context of rolling motion.
NAkid
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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?
 
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NAkid said:

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?
 
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