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Principle of Work/Energy Plane Motion (example continued)

Bar, cde, of length R is fastened to a ring of radius R as shown below where d denotes the

center of mass of the bar and c denotes the center of mass of the ring. Both the bar and ring

are of mass m. The bar-ring assembly is initially at rest when it starts to roll down the incline,

at angle β, with an angular speed, ω. Determine the kinetic energy of the assembly as a function

of m, R, and ω.

Strategy: The kinetic energy of a rigid body combines that for translation of the center of

mass and that for rotation about the center of mass. Kinetic energy is a scalar. So the total

kinetic energy is that for the bar plus that for the ring.

Solution: First consider the bar. Let v_d denote the speed of the center of mass of the bar

and ω the angular speed of the bar. The kinetic energy of the bar is then.

T_bar = (1/2) m v_d² + (1/2) I_zz^d ω²

I_zz^d is the mass moment of inertia of the bar about its center of mass and equals (1/12) mR²

From kinematics (for rolling) v_c = R ω Also v_d = v_c + ω k x ( ˗R/2 ) j

So v_d = R ω i + (R/2) ω i = (3R/2) ω i and v_d = (3R/2) ω

So the kinetic energy for the bar is T_bar = (1/2) m [ (3R/2)ω ]2 + (1/24) mR² ω² = (7/6) mR² ω²

Next consider the ring. T_ring = (1/2) m v_c² + (1/2) I_zz^c ω² where I_zz^c = mR²

T_ring = (1/2) m(Rω)² + (1/2) mR² ω² = mR² ω²

For the assembly of the bar and ring: T = (7/6) mR² ω² + mR² ω² = (13/6) mR² ω² (result)