C over these first 5 m at what average rate are you

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(c) Over these first 5 m, at what average rate are you doing work on the pulley by turning the crank? (d) At the instant the cylinder has moved 5 m, at what rate P crank are you doing work on the pulley by turning the crank? (e) Suppose you apply the same force to the crank as in part (a), but the cylinder slides frictionlessly up the ramp instead of rolling. Over the first 5 m of motion, do you exert a greater or smaller average power than in part (c)? Explain. (a) Applying the torque equation to the cylinder, we have f s r = 1 2 mr 2 a r = 1 2 ma Applying the force equation to the cylinder, we have T - mg sin - f s = ma T = ma + f s + mg sin = 3 2 ma + mg sin = 188 N 10
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Applying the torque equation to the pulley, we have F c R c - TR p = 1 2 M p R 2 p a R p = 1 2 M p aR p F c = 1 2 M p a + T R p R c = 385 N (b) The time taken to travel 5 m is d = 1 2 a Δ t 2 Δ t = r 2 d a = r (2)(5) 3 . 5 = 1 . 69 s During this time, we have v f = a Δ t = 5 . 92 m / s K f = 1 2 I ! 2 f + 1 2 mv 2 f = 1 2 1 2 + 1 mv 2 f = 3 4 mv 2 f P ave , cyl = K f - 0 Δ t = 3 mv 2 f 4 Δ t = 311 W (c) The average rate of work done by the crank is P ave , crank = F c R c Δ Δ t = F c R c ( d/R p ) Δ t = 711 W (d) At the instant the cylinder has traveled 5 m, the rate of work done by the crank is P crank = F c R c ! f = F c R c v f R p = 1422 W (e) If F c remains the same, but the cylinder slides, it has a faster acceleration up the ramp, since none of the work is diverted into rotational kinetic energy. Therefore, the cylinder takes less time to travel 5 m, though the work F c R c d/R p remains the same. Therefore, the average power of the crank is greater than in part (c). 11
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5. (33 pts) A 60-kg crate rests on a ramp inclined at 20 deg to the horizontal, sloping upward to the right. The crate has length 2 m parallel to the ramp, and length 1 m perpendicular to the ramp. You tie a copper rod to the center of the lower face of the crate (a perpendicular distance of 0.5 m from the surface of the ramp). Using the rod you pull horizontally to the left. (a) What minimum force must you exert on the rod in order for the crate to tip to the left, over its lower edge? (b) What minimum coe ffi cient of static friction between the crate and ramp is required in order to prevent the crate from slipping along the ramp when this force is applied? (c) Suppose the rod has a radius of 1 mm. If the rod has an initial length of 0.6 m, by what amount does it stretch when you pull on it with the force found in part (a)? Young’s modulus for copper is 11 10 10 Pa. (a) We use equilibrium of torques to find the required horizontal force F : 0 = X = Fh cos - Mgr ? , weight = Fh cos - Mg L 2 cos - W 2 sin F = Mg h cos L 2 cos - W 2 sin = (60)(9 . 8) (0 . 5) cos(20) 2 2 cos(20) - 1 2 sin(20) = 962 N (b) From equilibrium of forces parallel to the ramp, we have the following: 0 = μ s N - mg sin - F cos From equilibrium of forces perpendicular to the ramp, we have the following: 0 = N - mg cos + F sin 12
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Combining these equations, we can solve for the required minimum value of μ s : N = mg cos - F sin μ s N = μ s ( mg cos - F sin ) = mg sin + F cos μ s = mg sin + F cos mg cos - F sin = (60)(9 . 8) sin(20) + (962) cos(20) (60)(9 . 8) cos(20) - (962) sin(20) = 4 . 9 (c) From Hooke’s law, we have the following: F r 2 = Y Δ s s Δ s = sF Y r 2 = (0 . 6)(962) (11 10 10 )( )(0 . 001) 2 = 0 . 00167 m 13
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6. (33 pts) A satellite is in circular orbit around Earth, and requires 5 hours to complete 1 revolution (3600 s in 1 hour). Use the following: G = 6 . 67 10 - 11 N m 2 /kg 2 M E = 5 . 98 10 24 kg R E = 6 . 38 10 6 m (a) At what height above Earth’s surface is the satellite orbiting?
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