What is the tension force acting on the block?

[teatime]

1. A uniform sperical shell of mass Ms and radius R rotates about a vertical axis on frictionless bearings. A light cord passes around the equator of the shell over a pulley (a disk of mass Mp and radius r) and is attached to a small block of mass m that is otherwise free to fall under the influence of gravity.
A. What is the acceleration of the block?
B. What is the tension force acting on the block?
C. What is the tension force acting on the sphere?

Homework Equations

Isphere=(2/5) Ms R^2 Idisk=.5Mpr^2


3. Torque =rxf=Is(angular a)
RF=Is(angular a)
F=(2/5)MsR(ang. a)
Fnet=ma
Fs-mg=ma

 

[physixguru]

Draw an F.B.D..

 

[Moetasim]

A. The acceleration of the block can be found by using the equation Fnet=ma, where Fnet is the net force acting on the block and m is its mass. We can also use the equation Fs-mg=ma, where Fs is the tension force and mg is the force of gravity acting on the block. Since the block is otherwise free to fall, we can assume that the acceleration due to gravity (g) is acting on it. Therefore, the acceleration of the block can be expressed as a = (Fs-mg)/m.

B. The tension force acting on the block can be found by using the equation Fs-mg=ma. We know that the mass of the block is m and the acceleration is a, so we can rearrange the equation to solve for Fs. Therefore, the tension force acting on the block is Fs = m(a+g).

C. The tension force acting on the sphere can be found by using the equation F=(2/5)MsR(ang. a), where F is the net force acting on the sphere and ang. a is the angular acceleration. Since the sphere is rotating at a constant speed, the angular acceleration is 0. Therefore, the tension force acting on the sphere is F = 0. This makes sense because the only forces acting on the sphere are the normal force from the bearings and the tension force from the cord passing over the pulley, which cancel each other out.