This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Morgan, Mena Homework 10 Due: Nov 14 2007, 11:00 pm Inst: Opyrchal, H 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A simple pendulum consists of a small object of mass 1 . 6 kg hanging at the end of 1 . 7 m long light string that is connected to a pivot point. Find the magnitude of the torque (due to the force of gravity) about this pivot point when the string makes a 7 . 31047 angle with the vertical. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 3 . 39187 N m. Explanation: Let : m = 1 . 6 kg , L = 1 . 7 m , and = 7 . 31047 . 1 . 7 m 1 . 6 kg mg 1 . 7 m The torque is = F d = m g L sin = (1 . 6 kg)(9 . 8 m / s 2 )(1 . 7 m) sin7 . 31047 = 3 . 39187 N m . keywords: 002 (part 1 of 1) 10 points A uniform horizontal rod of mass 2 . 3 kg and length 0 . 75 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I = m 2 12 . . 75 m F 1 1 5 pivot 2 . 3 kg If 8 . 6 N force at an angle of 115 to the hor izontal acts on the rod as shown, what is the magnitude of the resulting angular accelera tion about the pivot point? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 6 . 04479 rad / s 2 . Explanation: Let : = 0 . 75 m , m = 2 . 3 kg , = 115 , and F = 8 . 6 N . By the parallel axis theorem, the moment of inertia of a stick pivoted at the end is I = I cm + m d 2 = 1 12 m 2 + m 2 2 = 1 3 m 2 . The sum of the torques (counterclockwise ro tation defined as positive) is X = F sin  m g 2 = I F sin  m g l 2 = 1 3 m 2 6 F sin  3 m g = 2 m l = 6 F sin  3 m g 2 m Since 6(8 . 6 N)sin(115 ) 3(2 . 3 kg)(9 . 8 m / s 2 ) = 20 . 8545 N then k ~ k = fl fl fl 20 . 8545 N fl fl fl 2(2 . 3 kg)(0 . 75 m) = 6 . 04479 rad / s 2 . Morgan, Mena Homework 10 Due: Nov 14 2007, 11:00 pm Inst: Opyrchal, H 2 keywords: 003 (part 1 of 1) 10 points An Atwood machine is constructed using a disk of mass 2 kg and radius 24 . 2 cm. 3 . 6 m 24 . 2 cm 2 kg 1 . 87 kg 1 . 32 kg What is the acceleration of the system? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 2864 m / s 2 . Explanation: Let : m = 2 kg , R = 24 . 2 cm = 0 . 242 m , m 1 = 1 . 32 kg , m 2 = 1 . 87 kg , h = 3 . 6 m . Consider the free body diagrams 1 . 87 kg 1 . 32 kg T 2 T 1 m 2 g m 1 g a a For the mass m 1 , F net = m 1 a = m 1 g T 1 T 1 = m 1 g m 1 a . (2) For the mass m 2 , F net = m 2 a = T 2 m 2 g T 2 = m 2 a + m 2 g . (3) The net acceleration a = r is in the direction of the heavier mass m 1 . The pulley is a solid disk, so I = 1 2 m r 2 ....
View
Full
Document
 Spring '12
 James
 Physics, Mass, Work, Light

Click to edit the document details