Your friend gets a flat tire while driving up a road with a 35.0degree slope. Wh

Question

Your friend gets a flat tire while driving up a road with a 35.0degree slope. While changing the tire he lets go of the spare and it begins to roll down the hill. His sister has stopped her car 4.51 m down the slope (of course this is very late on a cloudy night so she is there with her lights on). If we model the tire as a cylindrical shell with mass M and radius R that rolls without slipping. How fast (m/s) is the tire moving when it hits the sister's car? If instead we model the tire as a solid cylinder, how fast (m/s) would it be moving when it hits the sister's car?

Explanation / Answer

a) conservation of energy I = 2MR^2 1/2 I wf^2 = Mgh : I = 2MR^2 0.5*2* MR^2 (wf)^2 = Mg4.51*cos(35) wf = 6.01 /R Vf = wf *R = 6.01 m/sec b)I = 3/2 MR^2 0.5*1.5* MR^2 (wf)^2 = Mg4.51*cos(35) Vf = 6.94m/sec

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