GIVEN: M, a solid cylinder (M=2.31 kg, R=0.123 m) pivots on a thin, fixed, frict

Question

GIVEN: M, a solid cylinder (M=2.31 kg, R=0.123 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. Angular acceleration: 5.32x10^1 rad/s^2.

1) The force F in the previous question caused a torque 1 on the cylinder. The force F is removed and in its place an actual mass of m = 0.770 kg hung from the string and released. The weight of this mass also causes a torque 2 on the cylinder. T2 is _______T1 (greater than, less than or equal to)??

Explanation / Answer

when the cylinder was pulled down with a force F = 7.554 N

Tension is T = 7.554 N

and when a mass 0.77 kg is hanged,then tension is T = 0.77*9.8 = 7.554 N

in both the cases since the tension in the string is same then Torque acting in both cases is also same

so T2 is equal to T1

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