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qz6sol_3891f10

# qz6sol_3891f10 - 2 to the right and downward respectively...

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TA: Tomoyuki Nakayama Tuesday, October 12, 2010 PHY 2048: Physic 1, Discussion Section 3891 Quiz 6 (Homework Set #7) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ The figure below right shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block. The cart has mass m 1 = 0.800 kg and its center is initially at xy coordinates (-0.600 m, 0 m); the block has mass m 2 = 0.250 kg, and its center of mass is initially at xy coordinates (0 m, -0.200 m). a) In unit-vector notation what is the acceleration of the center of mass of the cart-block system? We take our positive directions for m 1 and m
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Unformatted text preview: 2 to the right and downward respectively. Newton’s 2 nd law yields 1: m 1 a = T, 2:m 2 a = m 2 g – T ⇒ a = m 2 g/(m 1 + m 2 ) = 2.33 m/s 2 In the coordinate system given in the figure, the accelerations of m 1 and m 2 are a 1 = +2.33 i , a 2 = -2.33 j The acceleration of the center of mass is a com = (m 1 a 1 + m 2 a 2 )/(m 1 + m 2 ) = 1.78(m/s 2 ) i – 0.555(m/s 2 ) j b) In unit-vector notation, what is the com of the system as a function of time t ? The velocity of the center of mass is v com = v com,0 + ∫ a com dt = 1.78t i – 0.555t j The center of mass of the system is initially located at x com,0 = (m 1 x 1,0 + m 2 x 2,0 )/(m 1 + m 2 )= -0.457 i – 0.0476 j The center of mass of the system is x com = x com,0 + ∫ v com dt = (-0.457 + 0.89t 2 )(m) i + (-0.0476 – 0.275t 2 )(m) j...
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