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Alonzo19
 3 years ago
M, a solid cylinder (M=2.11 kg, R=0.137 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 0.870 kg mass, i.e., F = 8.535 N.
A. Calculate the angular acceleration of the cylinder.
For A. I got this and it's correct "59.0 rad/s^2"
B. If instead of the force F an actual mass m = 0.870 kg is hung from the string, find the angular acceleration of the cylinder.
C.How far does m travel downward between 0.670 s and 0.870 s after the motion begins
Alonzo19
 3 years ago
M, a solid cylinder (M=2.11 kg, R=0.137 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 0.870 kg mass, i.e., F = 8.535 N. A. Calculate the angular acceleration of the cylinder. For A. I got this and it's correct "59.0 rad/s^2" B. If instead of the force F an actual mass m = 0.870 kg is hung from the string, find the angular acceleration of the cylinder. C.How far does m travel downward between 0.670 s and 0.870 s after the motion begins

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Alonzo19
 3 years ago
Best ResponseYou've already chosen the best response.0Here is the Image of the figure!

hba
 3 years ago
Best ResponseYou've already chosen the best response.1The force is equal to the weight of a hanging mass. This doesnt mean that there is actually a mass hanging off of it, it means there is a constant force applied of 8.142N. Once you see that the problem is simple. You have the equation T=I*alpha alpha is angular acceleration which is what you need so just solve for alpha. alpha=T/I where T is the torque applied and I is the moment of inertia. T = F*r (when torque is perpendicular) T = (8.142N) (0.133m) T= 1.083Nm Now find I, since it is a solid cylinder I=(1/2)Mr^2 I = (1/2) (1.99kg) (0.133m)^2 I = 0.0176 kg*m^2 now recall alpha = T/I alpha = (1.083Nm)/(.0176kg*m^2) ALPHA = 61.5 rad/s^2
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