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Home Q & A Board Science-Related Homework Help Physics
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inspector1994 inspector1994
wrote...
11 years ago
A 1.6  solid sphere (radius = 0.20 ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.55  high and 6.0  long. When the sphere reaches the bottom of the ramp, what is its rotational kinetic energy? When the sphere reaches the bottom of the ramp, what is its translational kinetic energy?
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wrote...
#1
11 years ago
PE = KE(rotational) + KE(translational)
=>mgh = 1/2 x I x omega^2 + 1/2mv^2
=>mgh = 1/2 x [2/5 x m x r^2] x [v/r]^2 + 1/2mv^2
=>gh = v^2[1/5 + 1/2]
=>v = sqrt[10/7 x gh]
=>v = sqrt[10 x 9.8 x 0.55/7]
=>v = 2.77 m/s
Thus KE(translational) = 1/2mv^2 = 1/2 x 1.6 x (2.77)^2 = 6.14 J
Thus KE(rotational) = 1/5mv^2 = 1/5 x 1.6 x (2.77)^2 = 2.46 J
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