phyhw17 - valdez (vv689) Homeowork 17 florin (58140) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: valdez (vv689) Homeowork 17 florin (58140) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A horizontal platform vibrates with simple harmonic motion in the horizontal direction with a period of 2 . 33 s. A body on the plat- form starts to slide when the amplitude of vibration reaches 0 . 334 m. Find the coefficient of static friction be- tween body and platform. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 247838. Explanation: Let : T = 2 . 33 s , A max = 0 . 334 m , and g = 9 . 8 m / s 2 . At each instant, there are two forces acting on the platform: the force responsible for the oscillation F = kx and the force of friction F s = N between the body and the platform. Applying Newtons second law horizontally, summationdisplay F x = kx + F s = ma platform The only force acting on the block in the horizontal direction is the frictional force, so F s = ma block If the block does not slide, its acceleration is the same as the platform: a block = a platform set = a The force of friction is F s = N = mg and from simple harmonic motion x = A cos t. a = d 2 x dt 2 = A 2 cos t, so the maximum acceleration is a max = A max 2 and the coefficient of static friction is = A max 2 g . Since T = 2 , = A max 4 2 g T 2 = (0 . 334 m)(4 2 ) (9 . 8 m / s 2 )(2 . 33 s) 2 = . 247838 . 002 (part 1 of 3) 10.0 points Consider a light rod of negligible mass and length L pivoted on a frictionless horizontal bearing at a point O . Attached to the end of the rod is a mass M 1 . Also, a second mass M 2 of equal size ( i.e. , M 1 = M 2 = M ) is attached to the rod parenleftbigg 2 5 L from the lower end parenrightbigg , as shown in the figure. 2 5 L L M 2 M 1 O What is the moment of inertia I about O ? 1. I = 17 16 M L 2 2. I = 13 9 M L 2 3. I = 73 64 M L 2 valdez (vv689) Homeowork 17 florin (58140) 2 4. I = 89 64 M L 2 5. I = 34 25 M L 2 correct 6. I = 74 49 M L 2 7. I = 58 49 M L 2 8. I = 25 16 M L 2 9. I = 10 9 M L 2 10. I = 29 25 M L 2 Explanation: The momentum of inertia is I M d 2 . There are two masses with I M 1 = M L 2 and I M 2 = M parenleftbigg L 2 5 L parenrightbigg 2 = M parenleftbigg 3 5 L parenrightbigg 2 = 9 25 M L 2 , so I = I M 1 + I M 2 = parenleftbigg 1 + 9 25 parenrightbigg M L 2 = 34 25 M L 2 . (1) 003 (part 2 of 3) 10.0 points If is in radians, what is Newtons second law of rotational motion for this pendulum? Use the small angle approximation. 1. I d 2 dt 2 = 9 7 M g L 2. I d 2 dt 2 = 5 3 M g L 3. I d 2 dt 2 = 6 5 M g L 4. I d 2 dt 2 = 12 9 M g L 5. I d 2 dt 2 = 3 2 M g L 6. I d 2 dt 2 = 12 7 M g L 7. I d 2 dt 2 = 16 9 M g L 8. I d 2 dt 2 = 9 5 M g L 9. I d 2 dt 2 = 8 5 M g L correct 10. I d 2 dt 2 = 4 3 M g L Explanation: Torque: r F...
View Full Document

Page1 / 8

phyhw17 - valdez (vv689) Homeowork 17 florin (58140) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online