Engr Help

Principle of Work/Energy Plane Motion (example continued)

Calculation of Work:

The work done includes that on the bar and that on the ring.

Note: The normal force and the frictional force between the incline and the ring contribute no

work. The normal force is perpendicular to the displacement, and for rolling, the frictional force contributes no work.

The only forces that contribute to work done on the bar/ring system are the weights of

the bar and of the ring.

The figure below shows the forces of gravity acting on the bar and on the ring.

Strategy: Both the mass center of the bar and the mass center of the ring (for one revolution)

displace 2πR down the incline from position 1 to position 2.

The work done by the gravity force on the bar is then (mg)( 2πR ) sin β.

The work done by the gravity force on the ring is also (mg)( 2πR ) sin β.

So the total work done on the bar/ring system is 4 mg π R sin β

Equate the work done to the change in KE: W_{1 ˗ 2}= T₂ - T₁

Therefore 4 mg π R sin β = (13/6) mR²ω²

ω² = (24/13) (g/R) π sin β and ω = √ [(24/13) (g/R) π sin β ] (result)