Question 6, chap 110, sect 5 part 1 of 1 10 points A uniform rod of mass 3.6 kg

Question

Question 6, chap 110, sect 5 part 1 of 1 10 points A uniform rod of mass 3.6 kg is 15 m long. The rod is pivoted about a horizontal, fric- tionless pin at the end of a thin extension (of negligible mass) a distance 15 m from the cen- ter of mass of the rod. Initially the rod makes an angle of 55° with the horizontal. The rod is released from rest at an angle of 55 with the horizontal, as shown in the figure below The acceleration of gravity is 9.8 m/s Hint: The moment of inertia of the rod about its center-of-mas is L- me. 12 3.6 kg 15 m 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

Let l = 15 m

     = 55 degree

    m =           3.6 kg

Rotational KE is   KR = 0.5*I*w^2

And gravitational KE is Ktrans = mgd

The inertia of the system is

I = ICM + md^2 = ml^2/12 + ml^2 = 13ml^2/12

Since the rod is uniform, its center of mass is located a distance l from the pivot. The vertical height of the center of mass above the horizontal is l*sin.

Using conservation of energy

Ki + Ui = Kf + Uf

Kf = Ui

0.5*I*w^2 = mgl* sin

13ml2w2/24 = mgl sin

w2 = 24*g* sin/13l

w = sqrt(24*g* sin/13l) = sqrt[(24*9.8*sin55)/(13*15)] = 0.9939936 rad/s

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