Posted by **Katie** on Friday, March 16, 2012 at 10:31pm.

The following objects are released from rest and allowed to roll down an inclined plane. Rank the objects by the order of which they will arrive at the bottom of the incline. (1 = the first to arrive, 2 = the second, etc.)

I = 2611 kg*m^2, M = 50 kg, R = 0.8 m

I = 2883 kg*m^2, M = 17 kg, R = 9.1 m

I = 4799 kg*m^2, M = 12 kg, R = 4.4 m

I = 2507 kg*m^2, M = 26 kg, R = 2.1 m

I = 1528 kg*m^2, M = 19 kg, R = 2.2 m

I = 4509 kg*m^2, M = 42 kg, R = 2 m

I = 1563 kg*m^2, M = 16 kg, R = 16.4 m

I = 2935 kg*m^2, M = 25 kg, R = 8.4 m

- physics -
**drwls**, Saturday, March 17, 2012 at 7:48am
They will arrive in the order of the final velocity, Vf. Conservation of energy will tell you what that is. Don't forget to include rotational kinetic energy.

M g H = (1/2) M Vf^2 + (1/2)*I*(Vf/R)^2

Vf^2 (M + I/R)^2 = 2 M g H

Vf^2 = 2 g H/[1 + I/(MR^2)]

The object with the lowest I/MR^2 will arrive first. Compare I/MR^2 for the various objects and rank them accordingly.

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