Kaplunovsky - practice 10

# Kaplunovsky - practice 10 - practice 10 – – Due: Apr 2...

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Unformatted text preview: practice 10 – – Due: Apr 2 2007, noon 1 Question 1, chap 12, sect 5. part 1 of 1 10 points Calculate the moment of inertia for b b b b b Axis 4 m 3 m m 2 m L L L L L Rods of length L are massless. 1. 32 mL 2 correct 2. 16 mL 2 3. 30 mL 2 4. 41 mL 2 5. 52 mL 2 Explanation: mL 2 + 2 m (2 L ) 2 + 3 m ( (2 L ) 2 + L 2 ) + 4 m ( L 2 + L 2 ) = mL 2 + 8 mL 2 + 3 m (5 L 2 ) + 4 m (2 L 2 ) = (1 + 8 + 15 + 8) mL 2 = 32 mL 2 Question 2, chap 12, sect 5. part 1 of 1 10 points The figure below shows a rigid system which can rotate. M = 1 . 1 kg, L = . 87 m, and the connecting rod as negligi- ble mass. Treat the masses as point particles. M x L 3 M What is the moment of inertia about an axis perpendicular to the paper and through the center of mass? Correct answer: 0 . 624443 kg m 2 (tolerance ± 1 %). Explanation: Basic Concept: I = summationdisplay m i r 2 i Solution: First find the center of mass. De- fine the origin to coincide with the left mass M . R = ∑ m i r i ∑ m i = 3 M L M + 3 M = 3 4 L. The moment of inertia of a system of point particles is given by I = summationdisplay m i r 2 i . Label the two moments of inertias as I left and I right . Remembering that the distances r i are with respect to the center of mass, we have I left = M parenleftbigg 3 4 L parenrightbigg 2 I right = 3 M parenleftbigg 1 4 L parenrightbigg 2 Hence I = I left + I right = M parenleftbigg 3 4 L parenrightbigg 2 + 3 M parenleftbigg 1 4 L parenrightbigg 2 = 0 . 624443 kg m 2 Question 3, chap 12, sect 5. part 1 of 2 10 points Consider a 2 kg square which has its mass only on its perimeter with each side a length 3 m. Hint: Use the parallel axis theorem and divide the square into parts. The mo- practice 10 – – Due: Apr 2 2007, noon 2 ment of inertia of a rod is I cm rod = 1 12 md 2 . d d What is the moment of inertia of the square about an axis perpendicular to the plane of the square at its center of mass? Correct answer: 6 kg m 2 (tolerance ± 1 %). Explanation: Basic Concept: The moment of inertia of a rod about its center of mass is found using I cm rod ≡ integraldisplay x 2 dm = integraldisplay + d 2 − d 2 x 2 m d dx = m d integraldisplay + d 2 − d 2 x 2 dx = m d x 3 3 vextendsingle vextendsingle vextendsingle vextendsingle + d 2 − d 2 = m 3 d bracketleftBigg parenleftbigg + d 2 parenrightbigg 3 − parenleftbigg − d 2 parenrightbigg 3 bracketrightBigg = 1 12 md 2 , where dm = λdx = m d dx . The parallel axis theorem is I = I cm + mℓ 2 . Solution: The moment of inertia of the square about its center of mass is I cm square = 1 3 md 2 . This can be shown by considering only one side of the square which is a rod whose mass is m 4 . The moment of inertia of a rod about its center of mass is I cm rod = 1 12 parenleftBig m 4 parenrightBig d 2 ....
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## 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.

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Kaplunovsky - practice 10 - practice 10 – – Due: Apr 2...

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