your friend gets a at tire while driving up a road with a 35.0 slope. While chan

Question

your friend gets a at tire while driving up a road with a 35.0 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 5.54 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, (a) How fast (m/s) is the tire moving when it hits the                             sister's car? (b) 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

the acceleration = g sin theta = 9.8 * sin 35 = 5.621 m/s^2

doing energy balance,

let velocity be v

then moment of inertia = MR^2

total energy initially = 5.54 sin 35 )*g*M = 31.14 M

total energy = 0.5*I*womega^2 + (0.5*mv^2) = 0.5*MR^2 * (V/R)^2 + (0.5*MV^2) = MV^2

so, 31.14 M = MV^2

V= velocity = 5.58 m/s

b)

doing energy balance,

let velocity be v

then moment of inertia = MR^2 )/2

total energy initially = 5.54 sin 35 )*g*M = 31.14 M

total energy = 0.5*I*womega^2 + (0.5*mv^2) = 0.5*((MR^2)/2) * (V/R)^2 + (0.5*MV^2) =0.75 MV^2

so, 31.14 M = 0.75MV^2

V= velocity = 6.4436 m/s

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