A uniform rod of mass 1.4 kg is 20 m long. The rod is pivoted about a horizontal

Question

A uniform rod of mass 1.4 kg is 20 m long. The rod is pivoted about a horizontal, fric- tionless pin at the end of a thin extension (of negligible mass) a distance 20 m from the cen- ter of mass of the rod. Initially the rod makes an angle of 60 with the horizontal. The rod is released from rest at an angle of 60 with the horizontal, as shown in the figure below The acceleration of gravity is 9.8 m/s2 . Hint: The moment of inertia of the rod about its center-of-mass is Icm = 1/12 m x l^2.

What is the angular speed of the rod at the instant the rod is in a horizontal position? Answer in units of rad/s.

Explanation / Answer

Here ,

let the angular speed of the rod is w

change in potential energy = change in kinetic energy

0.5 * I * w^2 = m * g * (L/2) * (sin(60))

0.5 * (1/12) * m * L^2 * w^2 = m * g * (L/2) * ( sin(60))

(1/6) * 20^2 * w^2 = 9.8 * (20/2) * (sin(60))

solving for w

w = 1.13 rad/s

the angular speed of the rod is 1.13 rad/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.