- #1
Punchlinegirl
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M, a solid cylinder (M=1.59 kg, R=0.127 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.750 kg mass, i.e., F = 7.357 N. Calculate the angular acceleration of the cylinder.
I used,
I*alpha= mgr
(1/2)mr^2 *alpha= mgr
(1/2)(1.59)(.127)^2 * alpha= (1.59)(9.8)(.127)
Solving for alpha gave me 155.9 rad/s^2
which wasn't right. Can someone help? Thanks.
I used,
I*alpha= mgr
(1/2)mr^2 *alpha= mgr
(1/2)(1.59)(.127)^2 * alpha= (1.59)(9.8)(.127)
Solving for alpha gave me 155.9 rad/s^2
which wasn't right. Can someone help? Thanks.