A 1.7 k g solid cylinder (radius = 0.20 m , length = 0.65 m ) is released from r

Question

A 1.7kg solid cylinder (radius = 0.20m , length = 0.65m ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.75m high and 5.0 m long.

Explanation / Answer

for rolling without slippng; v = w*R => v = 0.2*w moment of inertia, I = m*R^2/2 = 1.7*0.2^2/2 = 0.034 kg-m^2 from conservation of energy; initial gravitational P.E. = final translational K.E. + final rotational K.E. => mgh = 0.5*m*v^2 + 0.5*I*w^2 => 1.7*9.8*0.75 = 0.5*1.7*(0.2*w)^2 + 0.5*0.034*w^2 => w = 15.65 rad/s v = 0.2*w = 3.13 m/s total K.E. = 0.5*m*v^2 + 0.5*I*w^2 = 12.495 J rotational K.E. = 0.5*I*w^2 = 0.5*0.034*(15.65)^2 = 4.164 J translational K.E. = 0.5*m*v^2 = 0.5*1.7*3.13^2 = 8.33 J

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