A uniform rod (mass = 1.5 kg)is 2.0 m long.The rod is pivoted about a horizontal,frictionless pin through one end.The rod is released from rest in a horizontal position.What is the angular speed of the rod when the rod makes an angle of 30 $\degree$ with the horizontal? (The moment of inertia of the rod about the pin is 2.0 kg.m²).

Question

Quizplus · Accepted Answer

We can use conservation of energy to solve this problem. When the rod is released from rest, it has only potential energy, which is then converted into kinetic energy as the rod falls. At the final angle, all of the initial potential energy has been converted to rotational kinetic energy. We can write:

mgh = (1/2)Iω^2

where m is the mass of the rod, h is the height the center of mass falls, I is the moment of inertia about the pivot point, and ω is the angular velocity of the rod at the final angle.

In this problem, we can use the fact that the final angle is 30 degrees and the initial height is h = L(1 - cosθ), where L is the length of the rod and θ is the final angle. We can also use the fact that the moment of inertia of a uniform rod about one end is I = (1/3)ml^2.

Substituting in these values, we have:

mgL(1 - cosθ) = (1/2)(1/3)ml^2ω^2

Simplifying and solving for ω, we get:

ω = √(3g(1 - cosθ)/2L)

Plugging in the given values, we get:

ω = √(3(9.81)(1 - cos30)/4) ≈ 2.7 rad/s

Therefore, the answer is (C).

A Uniform Rod (Mass = 1 °\degree° With the Horizontal? (The Moment of Inertia of the Rod

A Uniform Rod (Mass = 1 $\degree$ With the Horizontal? (The Moment of Inertia of the Rod