In this problem, you will practice applying this formula to several situations i

Question

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar. You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.

Explanation / Answer

a = T/I, where T = torque and I = moment of inertia A) T = m1*g*L/2 - m2*g*L/2 = L/2*9*(m1 - m2) I = m1*(L/2)² + m2*(L/2)² = (m1 + m2)*(L/2)² a = L/2*g*(m1 - m2)/[L/2*(m1+ m2)*(L/2)²] = 2*[(m1 - m2)/(m1 + m2)]*g/L B) The moment of inertia changes. The moment of inertia of the bar is mbar*L²/12 so the total moment of inertia is mbar*L²/12 + (m1 + m2)*(L/2)² and a = L/2*g*(m1 - m2)/[L/2*(m1 + m2)*(L/2)² + mbar*L²/12] = 2*(m1 - m1)/(m1 + m2 + mbar/3)*g/L

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