M, a solid cylinder (M = 1.39 kg, R = 0.115 m) pivots on a thin, fixed, friction
ID: 1437859 • Letter: M
Question
M, a solid cylinder (M = 1.39 kg, R = 0.115 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 0.810 kg mass, i.e., F = 7.946 N. Calculate the angular acceleration of the cylinder. Tries 0/12 If Instead of the force F an actual mass m = 0.810 kg is hung from the string, Find the angular acceleration of the cylinder. Tries 0/12 How far does m travel downward between 0.390 s and 0.590 s after the motion begins? Tries 0/12 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 0.426 m in a time of 0.470 s. Find l_cm of the new cylinder. Tries 0/12Explanation / Answer
Torque = F*R= I where I =MR^2/2 and is the angualar accceleration.
F*R= MR^2/2
= 2F/(MR ) = 2*7.946/1.39/0.011 =1039rad/s^2
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When the force mg = 7.946is acting down, this force provides the linear
Acceleration' a 'of the mass and exerts a torque on M giving an angualar accceleration . a = R
For the mass M,
F = I /R = [MR^2/2] /R= {Ma/2}
For the mass m, ma = mg -F = mg - Ma/2
a = mg/ (M/2+m) = 7.946 / (0.855+0.8) = 4.80m/s^2
= a/R = 4.80 /0.0011 = 4363.6rad/s^2
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How far does mass m travel downward between t = 0.39 s and t = 0.59 s (Assuming motion begins at time t = 0.0 s)?
s = 0.5*a(t2^2 -t1^2) = 0.5*4.80*(0.59^2 - 0.39^2) =0.47m
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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 0.426 m in a time interval of 0.47 s. Find Icm of the new cylinder.
a = 2s/ (t^2) = 2*0.426/0.47*0.47 = 3.85 m/s^2
F = m g- ma = 7.96 - 0.8*3.85 = 4.88 N
F = I /R =Ia/R^2 = 4.88
I = 4.88*0.0011*0.0011/3.85 = 0.0015 kg m^2
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