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

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The acceleration of a solid sphere rolling down a 30-degree inclined plane is influenced by multiple forces, including gravity, friction, and the normal force. While the gravitational component along the incline is mg sin(θ), it is not the sole factor determining acceleration. To find the acceleration, both translational and rotational dynamics must be considered, applying Newton's second law. The concept of "rolling without slipping" indicates that the sphere completes one full revolution for every distance of 2πR traveled. Understanding these principles is crucial for accurately calculating the sphere's center of mass acceleration.
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A solid shere of mass M and radius R rolls without slipping down an inclined plane whose incline angle with the horizontal is 30 degrees. What is the acceleration of the sphere's center of mass?

gsinX is the acceleration, so it would be g/2 right?
 
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No. While mg \sin \theta is the component of the ball's weight along the incline, it is not the only force acting on the ball.
 
A solid shere of mass M and radius R rolls without slipping down an inclined plane whose incline angle with the horizontal is 30 degrees.
 
still not sure what you mean... so it doesn't slip, so what?
 
Start by identifying all the forces acting on the sphere. (Hint: I count 3)
 
force of friction, normal force, and force of gravity
 
Right. To solve for the acceleration, you'll need to apply Newton's 2nd law for both translation and rotation.
 
The point of "rolling without slipping" is that the sphere makes on complete revolution while going a distance 2\pi R
 
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