This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Niu (qn269) – HW07 – Tsoi – (58020) 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rod can pivot at one end and is free to rotate without friction about a vertical axis, as shown. A force vector F is applied at the other end, at an angle θ to the rod. L m F θ If vector F were to be applied perpendicular to the rod, at what distance d from the axis of rotation should it be applied in order to produce the same torque vector τ ? 1. d = √ 2 L 2. d = L tan θ 3. d = L 4. d = L sin θ correct 5. d = L cos θ Explanation: The torque the force generates is τ = F L sin θ . Thus the distance is L sin θ . 002 (part 1 of 4) 10.0 points Consider a rod of length L and mass m which is pivoted at one end. An object with mass m is attached to the free end of the rod. C L m m 24 ◦ Determine the moment of inertia of the system with respect to the pivot point. The acceleration of gravity g = 9 . 8 m / s 2 . Consider the mass at the end of the rod to be a point particle. 1. I = 3 2 m L 2 2. I = 13 12 m L 2 3. I = 5 3 m L 2 4. None of these. 5. I = 4 3 m L 2 correct 6. I = 5 4 m L 2 7. I = L 2 Explanation: The moment of inertia of the rod with re spect to the pivot point is I rod = 1 3 mL 2 , and the moment of inertia of the mass m with respect to the pivot point is I mass = mL 2 . Thus the moment of inertia of the system is I = I rod + I m = 1 3 mL 2 + mL 2 = 4 3 mL 2 . 003 (part 2 of 4) 10.0 points Niu (qn269) – HW07 – Tsoi – (58020) 2 The length C in the figure represents the lo cation of the centerofmass of the rod plus mass system. Determine the position of the center of mass from the pivot point. 1. C = 5 8 L 2. C = L 3. None of these 4. C = 3 4 L correct 5. C = 1 2 L 6. C = 7 8 L Explanation: The center of mass of the rod is 1 2 L , so the center of mass of the rodplusmass system is C = ( 1 2 suppressL + L ) m m + m = 1 4 L + 1 2 L = 3 4 L. 004 (part 3 of 4) 10.0 points The unit is released from rest in the horizontal position. What is the kinetic energy of the unit when the rod momentarily has a vertical orienta tion? 1. None of these 2. K = 2 mg L 3. K = 1 2 mg L 4. K = 5 2 mg L 5. K = mg L 6. K = 3 2 mg L correct Explanation: The potential energy released can be com puted in two ways: Δ U = Δ U vextendsingle vextendsingle vextendsingle cm of rod + Δ U vextendsingle vextendsingle vextendsingle mass = mg L 2 + mg L = 3 2 mg L, or Δ U  rod+m = Δ U  cm of rod+m = (2 m ) g parenleftbigg 3 4 L parenrightbigg = 3 2 mg L. 005 (part 4 of 4) 10.0 points Let the kinetic energy be K V at the vertical position and the moment of inertia of the system be I ....
View
Full
Document
This note was uploaded on 09/01/2009 for the course PHY M taught by Professor Staff during the Spring '09 term at University of Texas.
 Spring '09
 Staff
 Physics, Friction

Click to edit the document details