# 7AMT2answers - 1 It does not reach the top 1 5 h 6(energy...

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Unformatted text preview: 1. It does not reach the top. 1. 5 h 6 (energy conservation, circular motion, normal force = 0) 2. 2 3 q gh 3 ( ~v = q gh 3 (- cos θ ˆ i + sin θ ˆ j ) on circular path, where θ is the angle from the vertical, cos θ = 2 3 at break-o point, then projectile motion) 2. I actually meant the force was applied at one edge, as shown in the gure, in case that confused you. Trust the gure. Answer is Mg 2 (torque about bottom-right corner 6 = 0 ) 3. There's a minor typo - the speed of the raindrops should have been v r in all 3 parts, but it's okay if you used v in (2) and (3). Replace v r in the answers below by v to compare with your answers. Using 'D' for the density of water in order to not mix it with the 'd' in derivatives, A = πR 2 and M w = DAH = mass of the water in the barrel when it is full. 1. M M + M w v 2. Mv + M w v r cos θ M + M w (one nice way is by going into a reference frame where the rain seems to fall vertically and using the result from 1) 3....
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7AMT2answers - 1 It does not reach the top 1 5 h 6(energy...

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