A barbell is mounted on a nearly frictionless axle through its center. The low-m

Question

A barbell is mounted on a nearly frictionless axle through its center. The low-mass rod has a length d = 0.16 m, and each ball has a mass m = 0.4 kg. At this instant, there are two forces of equal magnitude F applied to the system as shown, with the directions indicated, and at this instant the angular velocity is 60 radians/s, counterclockwise, and the angle = 35 degrees. In the next 0.004 s, the angular momentum relative to the center increases by an amount 4 10-4 kg·m2/s. What is the magnitude of each force?

Explanation / Answer

torque is defined as follows:
torque = dL/dt
r*F*sin = dL/dt

[d/2]*F*sin = dL/dt

hence, the magnitude of each force is,

F = 2[dL/dt] / [r*sin] = 2*[ 4x10-4 / 0.004] / [0.16*sin35o] = 2.18 N

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