The thin uniform rod in the figure has length 4.0 m and can pivot about a horizo

Question

The thin uniform rod in the figure has length 4.0 m and can pivot about a horizontal, frictionless pin through one end. It is released from rest at angle ? = 50

A laser emits a narrow beam of light. The radius o 10^-3 m, and the power is 1.20 The thin uniform rod in the figure has length 4.0 m and can pivot about a horizontal, frictionless pin through one end. It is released from rest at angle ? = 50 degree above the horizontal. Use the principle of conservation of energy to determine the angular speed of the rod as it passes through the horizontal position. Assume free-fall acceleration to be equal to 9.83 m/s^2.

Explanation / Answer

Here ,

for the rod ,moment of inertia about the end is

I = ml^2/3

I = m*4^2/3

I = 16m/3 Kg.m^2

Using energy conservation ,

I*w = mgh

16*m/3 * w = m*9.8 * 4 * sin(50)

w = 5.63 rad/s

the angular speed of the rod as it passes through the horizontal position is 5.63 rad/s

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