CH-13 Solutions

# CH-13 Solutions - homework 10 BROWNING, ADAM Due: Nov 9...

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

Unformatted text preview: homework 10 BROWNING, ADAM Due: Nov 9 2007, 1:00 am 1 Question 1, chap 11, sect 3. part 1 of 1 10 points A uniform disk of radius 3 . 7 m and mass 1 . 2 kg is suspended from a pivot 0 . 518 m above its center of mass. The acceleration of gravity is 9 . 8 m / s 2 . axis Find the angular frequency for small os- cillations. Correct answer: 0 . 844776 rad / s (tolerance 1 %). Explanation: Basic Concepts The physical pendulum: = I = mg d sin = d 2 dt 2 so that the angular frequency for small oscil- lations (sin ) is = radicalbigg mg d I . Parallel axis theorem I = I + ma 2 Solution: We need the moment of inertia of the disk about the pivot point, which we call P. The moment of inertia of a uniform disk about its center is I disk = 1 2 mR 2 , but here the disk is rotating about P, a dis- tance d from the center of mass. The parallel axis theorem lets us move the axis of rotation a distance d : I P = 1 2 mR 2 + md 2 = m parenleftbigg R 2 2 + d 2 parenrightbigg . Then using the formula for the small angle oscillation frequency of a physical pendulum (see Basic Concepts above), we obtain = radicalBigg mg d I P = radicaltp radicalvertex radicalvertex radicalvertex radicalbt mg d m parenleftbigg R 2 2 + d 2 parenrightbigg or = radicaltp radicalvertex radicalvertex radicalvertex radicalbt g d R 2 2 + d 2 = radicaltp radicalvertex radicalvertex radicalvertex radicalbt (9 . 8 m / s 2 )(0 . 518 m) (3 . 7 m) 2 2 + (0 . 518 m) 2 = 0 . 844776 rad / s . Question 2, chap 11, sect 3. part 1 of 3 10 points Hint: The moment of inertia of a uniform rod about its center-of-mass is 1 12 M L 2 . Consider a uniform rod with a mass m and length L pivoted on a frictionless horizontal bearing at a point O parenleftbigg 4 7 L from the lower end parenrightbigg , as shown in the figure. 4 7 L L O The moment of inertia I of the rod about the pivot point O is given by homework 10 BROWNING, ADAM Due: Nov 9 2007, 1:00 am 2 1. I = 31 147 M L 2 2. I = 19 192 M L 2 3. I = 43 192 M L 2 4. I = 7 81 M L 2 5. I = 19 147 M L 2 6. I = 19 81 M L 2 7. I = 1 9 M L 2 8. I = 7 75 M L 2 9. I = 13 147 M L 2 correct 10. I = 7 36 M L 2 Explanation: Basic Concepts: Using the parallel axis theorem, the momentum of inertia is I M d 2 . In this case there are two masses with I O = I cm + M D 2 , where D = parenleftbigg 4 7 1 2 parenrightbigg L = 1 14 L, so (1) I O = M bracketleftBigg 1 12 + parenleftbigg 1 14 parenrightbigg 2 bracketrightBigg L 2 = 13 147 M L 2 . (2) Question 3, chap 11, sect 3. part 2 of 3 10 points Hint: Use the small angle approximation. If is in radians, then Newtons second law for rotational motion for this pendulum is given by 1....
View Full Document

## This note was uploaded on 02/24/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

### Page1 / 12

CH-13 Solutions - homework 10 BROWNING, ADAM Due: Nov 9...

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

View Full Document
Ask a homework question - tutors are online