As you approached the I-74 exit in your uncle\'s little red sports car convertib

Question

As you approached the I-74 exit in your uncle's little red sports car convertible you happened to notice a cat crossing the road. Not that you have any particular love of cats (especially this one whose snooty disposition seems to beckon for a game of chicken) you swerve out of the way spilling scalding hot coffee all over your new summer duds. Unknown to you at the time, the auto mechanic, having had an uneventful social life and being consequently somewhat incoherent, had inadvertently forgotten to fasten the lug nuts on one of your uncle's tires. Given your velocity on the exit ramp as 2.8 m/s, a tire mass of 10 kg and radius 30 cm: 1) What is the angular momentum of uncle's tire immediately upon its departure? (Assume the tire is a solid cylinder) in kg m2/s? Angular Momentum = kg m2/s 2) Next, the tire cruises up an embankment at 30

Explanation / Answer

help: KEi = 1/2mv^2 + 1/2I(v/r)^2 KEi = (1/2)(10kg)(2.8)^2 + (1/2)((1/2)(10kg)(0.3m)^2))(2.8m/s / 0.3m)^2 KEi = 58.8 m^2 kg / s^2 KEf = KEi - PE KEf = 58.8 m^2 kg / s^2 - (10kg)(-9.8 m/s ^2)(1sin30) KEf = 107.8 m^2 kg / s^2 KEf = 1/2mv^2 107.8 m^2 kg / s^2 = (1/2)(10kg)(v^2) v=4.64 m/s

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