What Is the Acceleration of a Rolling Sphere Down an Inclined Plane?

[UrbanXrisis]

a solid sphere of mass M and radius R rolls without slipping down an incluned plane whose incline angle with the horizontal is 30 degrees. That is the acceleration f the sphere's center of mass?

not sure how to attack this problem. Any equations and leads would be appreciated.

 

[Andrew Mason]

a solid sphere of mass M and radius R rolls without slipping down an incluned plane whose incline angle with the horizontal is 30 degrees. That is the acceleration f the sphere's center of mass?

not sure how to attack this problem. Any equations and leads would be appreciated.

This question was answered in your previous post at:

https://www.physicsforums.com/showthread.php?t=66391

The key is to think of the angular acceleration of the centre of mass about the point on the rim that is in momentary contact with the surface. While one does not think of the centre of mass as experiencing angular acceleration since it moves in a straight line, there is continuous angular acceleration about a pivot point that moves with the centre of mass.

AM

 

[thepatientmental]

To solve this problem, we can use the equations for rotational motion and conservation of energy.

First, we need to find the moment of inertia of the solid sphere about its center of mass. For a solid sphere, the moment of inertia is given by I = (2/5)MR^2.

Next, we can use the equation for the net torque on a rolling object, which is equal to the product of the moment of inertia and the angular acceleration. In this case, the net torque is caused by the force of gravity and the friction force between the sphere and the incline.

Since the sphere is rolling without slipping, we can also use the equation v = ωR, where v is the linear velocity of the center of mass and ω is the angular velocity.

Finally, we can use conservation of energy to relate the potential energy at the top of the incline to the kinetic energy at the bottom.

Putting all of this together, we can solve for the acceleration of the center of mass using the following equation:

a = (2/7)g sinθ

where g is the acceleration due to gravity and θ is the incline angle.

So in this case, the acceleration of the center of mass of the solid sphere would be approximately 0.34g, where g is the acceleration due to gravity. This means that the sphere will accelerate down the incline at a rate of 0.34 times the acceleration due to gravity.

I hope this helps! If you need more information or clarification, please let me know.