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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
 one year ago
 one year 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
 one year ago
 one year ago

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Alonzo19Best ResponseYou've already chosen the best response.0
Here is the Image of the figure!
 one year ago

hbaBest ResponseYou've already chosen the best response.1
The 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
 one year ago
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