# ps10sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01T Fall Term 2004 Problem Set 10: Torque, Rotational Dynamics, Physical Pendulum, Angular Momentum; Solutions Problem 1: ( Moment of Inertia ) A 1" US Standard Washer has inner radius r 1 = 1.35 × 10 2 m and an outer radius 3 r 2 = 3.10 × 10 2 m . The washer is approximately d = 4.0 × 10 m thick. The density of the washer 10 kg m 3 3 is ρ = 7.8 × . Calculate the moment of inertia of the washer about an axis passing 2 through the center of mass and show that it is equal to I = 1 mr 0 + r i 2 ) . w ( cm 2 1

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Problem 2: Experiment 09 Physical Pendulum Part One: Ruler Pendulum The ruler has a mass m r = 0.159 kg, a width a = 0.028 m, a length b = 1.00 m, and the distance l from the pivot point to the center of mass is = 0.479 m. Enter your measured period into the T meas column of the table below and calculate the other entries using the formulas / T ideal = 2 π l g and T / theory = 2 2 I 1 + θ 0 ml 2 16 with g = 9.805 ms -2 . Solution: The first part of the analysis of the experiment is to calculate the moment of inertia about an axis passing through the center of mass, perpendicular to the plane formed by the sides of the ruler. In particular, choose Cartesian coordinates with the origin at the center of mass, and the x-axis along the length, and y-axis along the width. The mass per unit area σ = mass m 1 = . Area ab 2 The mass element is rotating about the z-axis in a circular orbit with radius r = ( x + y 2 ) 12 , so the moment of inertia about the center of mass of the ruler is 3

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xb 2 y a 2 = = 2 2 I cm = ( r , cm ) d m = σ ( x + d d x 2 ) body x =− b 2 y a 2 We first do the integral in the y-direction, I = m 1 2 ⎜⎜ x y + y b 2 3 = = ⎛⎛ 2 3 ya 2 dx = m 1 2 a = x a + dx cm y a 2 3 ab x b 2 ⎝⎝ ab x b 2 1 2 We now do the integral in the x-direction 3 3 3 3 I S m 1 x a 2 a 2 = = a + x = m 1 b a + b = m 1 ( b + a 2 ) x b 2 ab 3 1 ab 1 1 1 I S 2 2 2 = ( 0.159 kg ) ( ( 1.000 m ) + ( 0.028 m ) ) = 1.326 × 10 kg m 2 12 Now use the parallel axis theorem to calculate the moment of inertia about the pivot point, 2 2 I S = m 1 ( b + a 2 ) + m l . 12 1 c Using the data for the ruler, the moment about the pivot point is 2 I S = 1.326 × 10 2 kg m 2 + ( 0.159 kg )( 0.479 m ) 2 = 4.97 × 10 kg m 2 0 θ T meas T theory T ideal 0.10 1.622 1.623 1.390 0.25 1.633 1.629 1.390 0.45 1.642 1.643 1.390 Part Two: Added Mass Consider the effect of a brass weight clipped to the ruler. The weight is shaped like a washer with an outer radius r o = 0.016 m and an inner radius r i = 0.002 m; it has a mass m w = 0.050 kg.
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ps10sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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