(a) M, a solid cylinder (M=2.07 kg, R=0.129 m) pivots on a thin, fixed, friction
ID: 2090930 • Letter: #
Question
(a) M, a solid cylinder (M=2.07 kg, R=0.129 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 0.770 kg mass, i.e., F = 7.554 N. Calculate the angular acceleration of the cylinder.
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(b)If instead of the force F an actual mass m = 0.770 kg is hung from the string, find the angular acceleration of the cylinder.
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(c)How far does m travel downward between 0.490 s and 0.690 s after the motion begins?
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(d) The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.506 m in a time of 0.570 s. Find Icm of the new cylinder.
Please HELP!!
I got 56.58 for part (a) somehow. Is that correct? & what units should the answer be in?
For all the other parts I am completely stumped!
Thank you in advance!!
Explanation / Answer
(a) By ? = FXr =>? = 6.573 x 0.131 =>? = 0.861 N-m By ? = I? =>? = ?/I =>? = 0.861/[1/2mr^2] =>? = 0.861/[1/2 x 2.19 x (0.131)^2] =>? = 45.82 rad/sec (b) Now F = mg =>F = 0.670 x 9.8 = 6.566 N By ? = FXr =>? = 6.566 x 0.131 =>? = 0.860 N-m By ? = I? =>? = ?/I =>? = 0.860/[1/2mr^2] =>? = 0.860/[1/2 x 2.19 x (0.131)^2] =>? = 45.77 rad/sec (c) By a = r? =>a = 0.131 x 45.77 = 6 m/s^2 thus by s = ut + 1/2at^2 =>s(0.630) = 0 + 1/2 x 6 x (0.630)^2 = 1.19 m =>s(0.830) = 0 + 1/2 x 6 x (0.830)^2 = 2.07 m =>s(0.830) - s(0.630) = 0.88 m (d) By s = ut + 1/2at^2 =>0.365 = 0 + 1/2 x a x (0.530)^2 =>a = 2.6 m/s^2 =>? = a/r = 2.6/0.131 = 19.84 rad/sec =>By ? = I? =>I = ?/? =>I = 0.860/19.84 = 43.35 x 10^-3 kg-m^2
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