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A homogeneous rod of length L=1.47 mL=1.47 m and mass m=4.43 kgm=4.43 kg is pivoted about...

A homogeneous rod of length L=1.47 mL=1.47 m and mass m=4.43 kgm=4.43 kg is pivoted about one end. It starts at an angle of θ=27.8 ∘θ=27.8 ∘ with the vertical as shown in Figure 1. You may find it useful to use I=13mL2I=13mL2 for a rod pivoted about one end. The rod has just been released from rest at θ.

1) What is the magnitude of the torque acting on the rod in this position?

2) What is the angular acceleration of the rod?

3) What is the acceleration of a point on the end of the rod?

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Homework Answers

Answer #1

Torque is product of perpendicular distance of the force form the pivot and the force.

By the relation of second law of circular motion, we can calculate angular acceleration.

And the resultant acceleration of the point at the end of the rod is the resultant of tangential and angular acceleration.

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