Kaplunovsky - practice 13

Kaplunovsky - practice 13 - practice 13 – – Due: Apr 23...

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: practice 13 – – Due: Apr 23 2007, noon 1 Question 1, chap 11, sect 3. part 1 of 1 5 points The equation of motion for a torsion pen- dulum (in small angle oscillation) is given by: summationdisplay τ : I α =- κθ, where α = d 2 θ dt 2 . What is its period of oscillation. (Hint: Using dimension analysis, you should be able to eliminate many of the incorrect choices.) 1. T = 2 π radicalbigg I κ correct 2. T = 2 π radicalbigg κ I 3. T = κ I 4. T = radicalbigg κ I 5. T = radicalbigg I κ 6. T = 2 π I κ 7. T = 2 π κ I 8. T = I κ Explanation: With α = d 2 θ dt 2 , the equation of motion is: τ = I α = I d 2 θ dt 2 =- κθ = ⇒ d 2 θ dt 2 =- κ I θ. Compare with the standard simple harmonic motion (SHM) equation d 2 θ dt 2 =- ω 2 θ, we find that ω = radicalbigg κ I . This, in turn, means that the period is given by: T = 2 π ω = 2 π radicalbigg I κ . Question 2, chap 15, sect 3. part 1 of 1 10 points A uniform plank of mass m is pivoted at one end. A spring of force constant k is attached to the center of the plank, as shown in the figure. The height of the pivot has been adjusted so that the plank will be in equilibrium when it is horizontally oriented. I CM rod = 1 12 mL 2 . L m θ k Find the period of small oscillation about the equilibrium point. 1. T = 2 π radicalbigg m k 2. T = 2 π radicalbigg m 3 k practice 13 – – Due: Apr 23 2007, noon 2 3. T = 2 π radicalbigg m 2 k 4. T = π radicalbigg 3 m 2 k 5. T = π radicalbigg m 3 k 6. T = π radicalbigg m 2 k 7. T = 2 π radicalbigg 2 m 3 k 8. T = 4 π radicalbigg m 3 k correct 9. T = π radicalbigg 2 m 3 k 10. T = 2 π radicalbigg 3 m 2 k Explanation: summationdisplay τ = I α Using the parallel axis theorem, the rotational inertia of the plank about the pivot point is I = 1 12 mL 2 + m parenleftbigg L 2 parenrightbigg 2 = 1 3 mL 2 . (1) From the free-body diagram, after displace- ment by a small angle θ , we have summationdisplay τ : parenleftbigg- mg L 2 + k y L 2 parenrightbigg cos θ = I α parenleftbigg- mg L 2 + k L sin θ L 4 parenrightbigg cos θ = I α, (2) where y = ( L/ 2) sin θ is the displacement from equilibrium. Using I from Eq. (1), and for small θ , cos θ = 1 and sin θ = θ , we have mg L 2- k L 2 θ 4 = 1 3 mL 2 d 2 θ dt 2 . (3) Substitute θ ′ = θ- 2 mg k L and d 2 θ ′ dt 2 = d 2 θ dt 2 , we have d 2 θ ′ dt 2 =- 3 k 4 m θ ′ (4) where the coefficient of θ ′ is ω 2 = 3 k 4 m . There- fore T ≡ 2 π ω = 2 π radicalbigg 4 m 3 k . (5) Alternate Solution: Use conservation of energy. E = mg y + 1 2 I ω 2 + 1 2 k y 2 = mg y + 1 2 parenleftbigg 1 3 mL 2 parenrightbigg ω 2 + 1 2 k y 2 = mg y + 2 3 mv 2 + 1 2 k y 2 Differentiate with respect to t ....
View Full Document

This note was uploaded on 01/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 12

Kaplunovsky - practice 13 - practice 13 – – Due: Apr 23...

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