{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2-8 (9) - Create assignment 00223 Homework 8 Oct 24 at 7:27...

This preview shows pages 1–2. Sign up to view the full content.

Create assignment, 00223, Homework 8, Oct 24 at 7:27 pm 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Cable Over a Pulley 10:07, trigonometry, numeric, > 1 min, nor- mal. 001 A cable passes over a pulley. Because the cable grips the pulley and the pulley has non- zero mass, the tension in the cable is not the same on opposite sides of the pulley. The force on one side is 120 N, and the force on the other side is 100 N. Assuming that the pulley is a uniform disk of mass 2 . 1 kg and ra- dius 0 . 81 m, determine the magnitude of its angular acceleration. Correct answer: 23 . 5156 rad / s 2 . Explanation: The resultant torque is given by (120 N)(0 . 81 m) - (100 N)(0 . 81 m) = 16 . 2 N m The moment of inertia is: I = 1 2 m r 2 = 1 2 (2 . 1 kg)(0 . 81 m) 2 = 0 . 688905 kg m 2 . Then, τ = I α gives α = τ I = 16 . 2 N m 0 . 688905 kg m 2 = 23 . 5156 rad / s 2 . Complex Atwood Machine 10:07, trigonometry, numeric, > 1 min, nor- mal. 002 An Atwood machine is constructed using two wheels (with the masses concentrated at the rims). The left wheel has a mass of 2 kg and radius 22 cm. The right wheel has a mass of 2 . 5 kg and radius 30 cm. The acceleration of gravity is 9 . 8 m / s 2 . 3 m 1 m 4 m 2 m The hanging mass on the left is 1 . 5 kg and on the right 1 kg. What is the acceleration of the system? Correct answer: 0 . 7 m / s 2 . Explanation: Basic Concepts: The net acceleration a = is in the direction of the heavier mass m 1 . Each pulley’s mass is concentrated on the rim, so I = m pulley r 2 3 m 1 1 m 4 m 2 2 2 m T 1 T 3 T a a r r For each pulley, τ net = = mr 2 a r · = mra so that for the leftmost pulley, T 1 r 1 - T 2 r 1 = m 1 r 1 a m 1 a = T 1 - T 2 (1) and for the rightmost pulley, T 2 r 2 - T 3 r 2 = m 2 r 2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern