week9 solution - Physics 1A Mechanics Winter 2016...

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Physics 1A: Mechanics Winter 2016 Discussion: Week 9 Agnieszka Wergieluk Graduate TA [email protected] O ffi ce hours: Mondays, Tuesdays, Wednesdays 4-7 p.m. by appointment! (Alternatively, just e-mail me your questions.) Problem 1 Calculate the moment of inertia of half a uniform solid cylinder of total mass M , radius R and height L , rotating about the z -axis as shown on the figure below. From the definition, I = Z dm r 2 . (1) Now, dm = dV , (2) where = M V = 2 M R 2 L (3) and dV = dr r d dz . (4) Plugging that into (1) and applying the correct integral limits, we have now I = 2 M R 2 L Z R 0 dr r 3 Z 0 d Z L 0 dz = 2 M R 2 r 4 4 R 0 = MR 2 2 . (5) 1
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Agnieszka Wergieluk, Physics 1A Discussion: Week 9 - Winter 2016 Problem 2 A massless rod has two masses - which you can treat as point-like - attached to it, as shown in the picture. Initially, the rod is aligned with the vertical. Then it is deviated from the vertical by an infinitesimal angle. What happens next? (Your answer should be both qualitative and quantitative.) What is the acceleration of the rod? What is the angular velocity at = 2 ? What is the linear velocity of the masses? Let us consider what are the forces at play by making an appropriate drawing: 2
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Agnieszka Wergieluk, Physics 1A Discussion: Week 9 - Winter 2016 We know that the whole system - the rod with the masses on each end - will not move as a whole (i.e., its center of mass won’t move) because it is pivoted. You can look at it the following way: the components of forces of gravity, m 1 g and m 2 g , along the rod will not overcome the reaction force of the pivot, and thus the system as a whole will not move. What about the components of m 1 g and m 2 g perpendicular to the rod? These will act to rotate the rod around the pivot! The component m 1 g sin will act to rotate the rod clockwise, and the component m 2 g sin will act to rotate the rod counterclockwise. If we take the clockwise direction to be positive (it’s very important to choose the direction and then stick to it! ), then the total torque acting on the rod is = L 1 m 1 g sin - L 2 m 2 g sin = ( L 1 m 1 - L 2 m 2 ) g sin . (6) What is going to happen now fully depends on the sign of the term in the bracket. If L 1 m 1 - L 2 m 2 < 0 , (7) then the torque acts counterclockwise, meaning it acts to put the rod back towards its equilibrium position. If, on the other hand, L 1 m 1 - L 2 m 2 > 0 , (8) then the torque acts clockwise and this means that the rod will continue to rotate. Think about the relative magnitudes if m 1 , m 1 , L 1 and L 2 that will produce both of those e ects. We can then calculate what is the acceleration of the rod, using = I ) = I . (9) The moment of inertia of this rod is I = X i m i r 2 i = m 1 L 2 1 + m 2 L 2 , (10) so that = ( L 1 m 1 - L 2 m 2 ) g sin m 1 L 2 1 + m 2 L 2 . (11) Now, let us note that = d !
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