Question

A uniform bar, of mass M, with seven evenly spaced holes is held by sliding the bar over a horizontal peg through one of the seven holes. The peg passes through hole C, and a cylinder hangs from a hook placed through hole B as shown above. The mass of the bar is equal to the mass of the cylinder, and the location of the center of mass of the bar is at the center of hole D. In this configuration, the bar-cylinder system remains motionless but is free to rotate around the peg in hole C. Frictional forces acting on the bar are negligible. In a clear, coherent paragraph-length response that may also contain equations, explain why the bar does not rotate in this configuration.

Please help!

Answer 1

It should be noted that because the torques caused by the bar and cylinder weights are equal, the bar-cylinder system remains static and does not rotate around the peg in hole C.

Because it is in equilibrium, the bar-cylinder system does not spin around the peg in hole C. The torque on the system caused by the cylinder hanging from hole B is balanced by the torque caused by the bar and its distribution around the center of mass at hole D.

The center of mass is positioned so that the torques generated by the weights of the bar and the cylinder are equal and opposite, resulting in a net torque on the system.