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Homework 24 - homework 24 BAUTISTA ALDO Due 4:00 am Version...

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homework 24 – BAUTISTA, ALDO – Due: Mar 29 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 3. 10 points. Three particles of mass 8 kg, 5 kg, and 8 kg are connected by rigid rods of negligible mass lying along the y axis and are placed at 5 m, - 3 m, and - 7 m , respectively as in the figure. The system rotates about the x axis with an angular speed of 2 . 5 rad / s . Contrary to what is observed in the figure, consider the masses to be point particles. 5 m - 3 m - 7 m 2 . 5 rad / s x 8 kg 5 kg 8 kg Find the moment of inertia about the x axis. Correct answer: 637 kg m 2 (tolerance ± 1 %). Explanation: Let : m 1 = 8 kg , m 2 = 5 kg , m 3 = 8 kg , y 1 = 5 m , y 2 = - 3 m , y 3 = - 7 m , and ω = 2 . 5 rad / s . The total rotational inertia of the system about the x axis is I = X m i r 2 i = 637 kg m 2 , where, r i = | y i | . Question 2 Part 2 of 3. 10 points. Find the total rotational energy of the sys- tem. Correct answer: 1990 . 62 J (tolerance ± 1 %). Explanation: Since ω = 2 . 5 rad / s , the total rotational energy is E = 1 2 I ω 2 = 1 2 (637 kg m 2 ) (2 . 5 rad / s) 2 = 1990 . 62 J . Question 3 Part 3 of 3. 10 points. Find the linear speed of the top particle of mass 8 kg in the figure. Correct answer: 12 . 5 m / s (tolerance ± 1 %). Explanation: The linear speed of each particles are v i = r i ω v 1 = r 1 ω = (5 m) (2 . 5 rad / s) = 12 . 5 m / s . Question 4 Part 1 of 1. 10 points. A 1900 kg car is being lowered by a winch. At the monent shown in the figure, the gear- box of the winch breaks and the car falls from rest. During the car’s fall, there is no slipping between the (massless) rope, the pulley, and the winch drum. The moment of inertia of the winch drum is 257 kg · m 2 and that of the pulley is 14 kg · m 2 . The radius of the drum is 86 cm and the radius of the pulley is 9 cm. The acceleration due to gravity is 9 . 81 m / s 2 .
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homework 24 – BAUTISTA, ALDO – Due: Mar 29 2006, 4:00 am 2 1900 kg 86 cm 9 m Find the speed of the car as it hits the ground. Correct answer: 9 . 18609 kN (tolerance ± 1 %). Explanation: Let : I d = 257 kg · m 2 , I p = 14 kg · m 2 , r d = 86 cm = 0 . 86 m , r p = 9 cm = 0 . 09 m , m b = 1900 kg , and Δ h = 9 m . Applying conservation of mechanical en- ergy ( K i = U f = 0) and v = r ω , 1 2 m v 2 + 1 2 I d ω d + 1 2 I p ω p - m g Δ h = 0 1 2 m v 2 + 1 2 I d v 2 r 2 d + 1 2 I p v 2 r 2 p - m g Δ h = 0 v = v u u u t 2 m g Δ h m + I d r 2 d + I p r 2 p = v u u u t 2 (1900 kg) (9 . 81 m / s 2 ) (9 m) 1900 kg + 257 kg · m 2 (0 . 86 m) 2 + 14 kg · m 2 (0 . 09 m) 2 = 9 . 18609 m / s .
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