Assignment 8 - PHYSICS 8A: PROBLEM SET 8 1. Angular Motion...

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PHYSICS 8A: PROBLEM SET 8 1. Angular Motion with Constant Acceleration (1.1) θ ( t )= θ 0 + ω 0 t + 1 2 αt 2 (A) In (1.1), θ is a funciton of time. (B) In (1.1), θ 0 is a constant. (C) In (1.1), ω 0 is a constant. (D) In (1.1), ω is a funciton of time. (E) ω 2 = ω 2 0 +2 α ( θ - θ 0 ) does not exhibit explicit t -dependence. (F) (1.2) ω = ω 0 + αt In (1.2), t is the time since the angular velocity was ω 0 . (G) Particle A ’s angular trajectory is given by (1.3) θ A ( t )= θ 0 + ω 0 t + 1 2 αt 2 We also know that particle B is also undergoing constant acceleration, and that at time t 1 , it has equal angular position, half the angular velocity, and twice the angular accerlation, that A did at time t = 0. We could ±gure out θ B ( t ) through a lot of calculus and algebra, or we can de±ne T = t - t 1 . Treating T like “time,” we quickling conclude that B ± s “initial” angle is θ 0 , its “initial” angular velocity is 1 2 ω 0 , and its acceleration is 2 α . Thus θ B = θ 0 + 1 2 ω 0 T + αT 2 . Substituting back in our de±nition of T , we see that θ B ( t )= θ 0 + 1 2 ω 0 ( t - t 1 )+ α ( t - t 1 ) 2 . (H) Angular velocity being the time-derivative of angle, we set B dt ( t )= A dt ( t ) 1 2 ω 0 +2 α ( t - t 1 )= ω 0 + αt , by di²erentiating our answer to (G). Thus we must have t = ω 0 2 α +2 t 1 , which is ω 0 2 α + t 1 after t 1 . 2. Torque and Angular Acceleration (A) We can ±gure out all the torques, so we can use = τ to ±nd α . We de±ne (+) to be counterclockwise both because m 1 >m 2 and because this is the most common convention (though the problem does not specify this). The magnitude of the torque τ 1 from m 1 is m 1 g l 2 since the force it exerts is perpendicular to its lever arm, which is half the length of the seesaw. τ 1 is positive because it would spin the seesaw counterclockwise, were it the only torque. Similarly, τ 2 = - m 2 g l 2 , since it would spin the seesaw clockwise. Thus τ = τ 1 + τ 2 = gl 2 ( m 1 - m 2 ). Since the mass of the seesaw itself is stated to be negligible, I = I 1 + I 2 = m 1 ( l 2 ) 2 + m 2 ( l 2 ) 2 . Thus we arrive at α = τ I = 2 g ( m 1 - m 2 ) l ( m 1 + m 2 ) . (B) Since m 1 >m 2 , α > 0 and by our convention, this corresponds to a counterclockwise rotation. (Had we used the opposite convention, we would have gotten α ( m 2 - m 1 ) instead, giving α < 0 and still ended up with counterclockwise.) Date : October 23, 2009. 1
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2 PHYSICS 8A: PROBLEM SET 8 (C) Now we are told to stop neglecting m bar . A rod rotating about its axis has I bar = 1 12 m bar l 2 = 1 3 m bar ( l 2 ) 2 . Thus the total rotational of the seesaw is now I = l 2 4 ( m 1 + m 2 + m bar 3 ) . The total torque is the same as in (A) because m bar g applies at 0 distance from the pivot, so we now Fnd that α = 6 g ( m 1 - m 2 ) l (3 m 1 +3 m 2 + m bar ) . (D) Changing
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Assignment 8 - PHYSICS 8A: PROBLEM SET 8 1. Angular Motion...

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