A 1.9 k g solid sphere (radius = 0.15 m ) is released from rest at the top of a

Question

A 1.9kg solid sphere (radius = 0.15m ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.65m high and 5.2m long.

Explanation / Answer

for rolling without slippng; v = w*R => v = 0.15*w moment of inertia, I = 2m*R^2/5 = 2*1.9*0.15^2/5 = 0.0171 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 => 1.9*9.8*0.65 = 0.5*1.9*(0.15*w)^2 + 0.5*0.0171*w^2 => w = 20.11 rad/s v = 0.15*w = 3.02 m/s total K.E. = 0.5*m*v^2 + 0.5*I*w^2 = 12.103 J rotational K.E. = 0.5*I*w^2 = 0.5*0.0171*(20.11)^2 = 3.46 J translational K.E. = 0.5*m*v^2 = 0.5*1.9*3.02^2 = 8.643 J

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