Part 1 A long uniform rod of length 4.29 m and mass 6.88 kg is pivoted about a h

Question

Part 1

A long uniform rod of length 4.29 m and
mass 6.88 kg is pivoted about a horizontal,
frictionless pin through one end. The rod is
released from rest in a vertical position.At the instant the rod is horizontal, ?nd its
angular speed. The acceleration of gravity is
9.8 m/s
2
.
Answer in units of rad/s

Part 2

At the same instant ?nd the magnitude of its
angular acceleration.
Answer in units of rad/s^2

Part 3

At the same instant ?nd the magnitude of the
acceleration of its center of mass.
Answer in units of m/s
2

Part 4


Finally, ?nd the magnitude of the reaction
force at the pivot (still at the same moment).
Answer in units of N

Explanation / Answer

Part 1 is an energy question.

Change in potential energy = m g * (1/2)L

Change in KE = (1/2) I w^2 = (1/2) * (1/3) m L^2 w^2

lost PE = gained KE so

    m g (1/2) L = (1/6) m L^2 w^2

w^2 = 3 g / L = 3 * 9.8 / 4.29 = 6.90

w = 2.627 rad/s

Part 2

This is a torque problem.

   total torque = I alpha

     (1/2) m g L = (1/3) m L^2 alpha

alpha = (3/2) g / L = (3/2) * 9.8 / 4.29 = 3.427 rad/s^2

Part 3

linear acc = alpha r = alpha * (1/2) L = 3.427 * (1/2) * 4.29 =

           = 7.35 m/s^2

Part 4

   total force = ma

    mg - force from pivot = ma

    force from pivot = mg - ma = m (g-a) =

           = 6.88 * (9.80 - 7.35) = 16.856 N

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