Rotational motion of a solid sphere of mass

[kaka01]

A solid sphere of mass m is released from rest from the rim of a hemispherical cup so that it rolls along the surface. If the rim of the hemisphere is kept horizontal, find the normal force exerted by the cup on the ball when the ball reaches the bottom of the cup.

 

[mikeph]

This sounds suspiciously like a homework question... anyway, the bottom of the cup is horizontal so the normal force must be equal and opposite to the weight of the ball, ie. acting upwards and of magnitude m*g.

 

[Doc Al]

This sounds suspiciously like a homework question...

Yep. It belongs in the Intro Physics forum.

anyway, the bottom of the cup is horizontal

No, the rim of the cup is kept horizontal.

 

[HallsofIvy]

Okay, so the tangent plane to the bottom of the cup is horizontal?

 

[PJC01]

As the solid sphere rolls along the surface of the hemispherical cup, it experiences both translational and rotational motion. The translational motion is caused by the force of gravity pulling the sphere towards the bottom of the cup, while the rotational motion is caused by the torque exerted on the sphere by the cup.

At the point of release, the sphere is at rest and the only force acting on it is the force of gravity. As it rolls down the cup, the force of gravity remains constant, but the normal force exerted by the cup increases in order to balance the torque exerted by the cup on the sphere.

When the sphere reaches the bottom of the cup, the normal force exerted by the cup on the sphere will be equal to the force of gravity acting on the sphere. This is because the sphere has reached its maximum speed and is no longer accelerating, meaning that the net force on the sphere is zero.

In order to determine the normal force at the bottom of the cup, we can use the equation for torque: τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. Since the sphere is rolling without slipping, we know that the angular acceleration is equal to the linear acceleration divided by the radius of the sphere, or α = a/r.

Substituting this into the torque equation, we get τ = I(a/r). We can also use the equation for rotational kinetic energy: K = 1/2Iω^2, where K is the kinetic energy, I is the moment of inertia, and ω is the angular velocity. Since the sphere is rolling without slipping, we know that the linear velocity is equal to the angular velocity multiplied by the radius of the sphere, or v = ωr.

Substituting this into the kinetic energy equation, we get K = 1/2I(v^2/r^2). Since the sphere is released from rest, its initial kinetic energy is zero, and at the bottom of the cup, its kinetic energy is equal to its final kinetic energy, which is equal to 1/2mv^2.

Setting these two equations equal to each other and solving for the normal force, we get:

1/2I(v^2/r^2) = 1/2mv^2

Solving for v^2, we get v^2 = 2gr, where g is the acceleration due to gravity