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

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SUMMARY

The maximum angle at which a solid cylinder rolls without slipping on an inclined plane, given that a block begins to slide at an angle of 11.86 degrees, can be determined using the equations of motion and torque. The relevant equations include mgsin(theta) + friction = ma and torque = friction*R = I(angular acceleration), where I = 0.5MR^2. By substituting these equations, the relationship between linear acceleration and angular acceleration can be established, leading to the calculation of the angle theta using the inverse sine function.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with rotational dynamics and torque
  • Knowledge of free body diagrams
  • Basic trigonometry, specifically inverse sine functions
NEXT STEPS
  • Study the relationship between linear velocity and angular velocity in rotational motion
  • Explore the concepts of static and kinetic friction in inclined planes
  • Learn about the moment of inertia for various shapes, particularly cylinders
  • Investigate the effects of different materials on friction and rolling motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for practical examples of rotational motion and friction on inclined planes.

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 [itex]v[/itex] and [itex]\omega[/itex]?
 

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