solution_pdf

# solution_pdf - tokarski(at23678 Torque Rolling...

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

tokarski (at23678) – Torque Rolling – Morozova – (11104) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A uniform rod has length 46 m and mass 9 kg . A mass of 9 kg is attached at one end. The other end of the rod is pivoted about the horizontal axis at O . 9 kg 9 kg 46 m 9 kg 9 kg pivot O 57 Determine the torque about O immediately after the rod-plus-mass system is released from the horizontal position. The accelera- tion of gravity is 9 . 8 m / s 2 . Correct answer: 6085 . 8 kg · m 2 / s 2 . Explanation: Let : m = m = 9 kg , L = 46 m , and θ = 57 . There are two forces on the system (the weight of the rod which can be calculated at the center of the rod, and the weight of the mass, so the torque is given by τ = m g parenleftbigg L + 1 2 L parenrightbigg = parenleftbigg 3 2 parenrightbigg m g L = parenleftbigg 3 2 parenrightbigg (9 kg) (9 . 8 m / s 2 ) (46 m) = 6085 . 8 kg · m 2 / s 2 . 002 (part 2 of 3) 10.0 points After the rod-plus-mass system is released, it rotates freely about the point O . Determine its angular velocity ω as the rod passes through the vertical direction. Correct answer: 0 . 692349 s 1 . Explanation: The moment of inertia of the system about the pivot is given by I = 1 3 m L 2 + m L 2 = 4 3 m L 2 . The center of mass of the system drops a height of 3 4 L as it passes through the vertical direction. By equating the gain of rotational energy and the loss of potential energy, we have 1 2 parenleftbigg 4 3 m L 2 parenrightbigg ω 2 = (2 m ) 3 4 g L ω = radicalbigg 9 g 4 L = radicalBigg 9 (9 . 8 m / s 2 ) 4 (46 m) = 0 . 692349 s 1 .

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