Can you help me on the problem and explain. Harmonie Oscillator. A uniform thin

Question

Can you help me on the problem and explain.

Harmonie Oscillator. A uniform thin rod of length e = 1.2 m and mass M = 3.0 kg is attached to a horizontal pivot ( middot ) at distance e/4 = 0.30 m below the top of the rod as shown. A cylinder of radius r = 2.0 cm and mass m = 1.8 kg is attached to the rod at distance e/4 = 0.30 m above the bottom of the rod. The rod-cylinder combination is pulled back so that the rod makes an angle of 15degree to the vertical as shown, and released. Ignoring friction and air drag, which choice best represents the period of oscillation of this pendulum?

Explanation / Answer

T = 2pi * sqrt ( I / mg*Lcm )

I = moment of inertia about the pivot.

I = moment of inertia of cyliner + moment of inertiia of rod about pivot.

I = (m*r^2/2 + mL^2/4 )   + (ML^2/12 + ML^2/16)

I = (1.8 * 2^2 * 10^-4/2 + 1.8*1.2^2/4 ) + (3*1.2^2/12 + 3*1.2^2/16)

I = 1.27836 Kg-m^2

Lcm = 1.8* 0.9 + 3*0.3   / (1.8+3) = 0.525 m from the pivot.

so T = 2pi* sqrt( 1.27836 / (4.8*9.8*0.525)) = 1.53 seconds... == 1.6 s

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