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Unformatted text preview: oldmidterm 03 KIM, JI Due: Oct 17 2007, noon 1 Question 1, chap 13, sect 1. part 1 of 1 10 points A constant torque of 28 . 3 N m is applied to a grindstone whose moment of inertia is . 11 kg m 2 . Using energy principles, and neglecting fric tion, find the angular speed after the grind stone has made 16 . 4 rev, assuming it started from rest. Correct answer: 36 . 6475 rev / s (tolerance 1 %). Explanation: Given : = 28 . 3 N m , I = 0 . 11 kg m 2 , and = 16 . 4 rev . The work done on the grindstone is W net = vector F vectors = F ( r ) = ( F r ) = Thus, W net = 1 2 I w 2 f 1 2 I w 2 i so = 1 2 I 2 and = radicalbigg 2 I = radicalBigg 2 (16 . 4 rev)(28 . 3 N m) . 11 kg m 2 2 rev rev 2 = 36 . 6475 rev / s . Question 2, chap 13, sect 2. part 1 of 1 10 points A wooden cylinder (in the form of a thin disk) of uniform density and a steel hoop are set side by side, released from rest at the same moment, and roll down an inclined plane towards a wall at the bottom. Hint: There is no energy loss due to friction. The cylinder has a larger radius than the hoop, but the hoop weighs more than the cylinder. Select the correct statement. 1. The hoop reaches the bottom first because I cylinder M cylinder R 2 cylinder > I hoop M hoop R 2 hoop . 2. The hoop and cylinder reach the bottom at the same time because friction is negligi ble. 3. The cylinder reaches the bottom first because I hoop M hoop R 2 hoop > I cylinder M cylinder R 2 cylinder . correct 4. The answer cannot be determined unless the radii and mass are equal. 5. The cylinder reaches the bottom first be cause its moment of inertia I cylinder > I hoop . 6. The hoop reaches the bottom first because I hoop M hoop R 2 hoop > I cylinder M cylinder R 2 cylinder . 7. The cylinder reaches the bottom first be cause its moment of inertia I hoop > I cylinder . 8. The hoop reaches the bottom first because its moment of inertia I hoop > I cylinder . 9. The cylinder reaches the bottom first be cause I cylinder M cylinder R 2 cylinder > I hoop M hoop R 2 hoop . 10. The hoop reaches the bottom first because its moment of inertia I cylinder > I hoop . Explanation: Say the two objects start at the top of an incline of vertical height h . Conservation of energy gives mg h = 1 2 mv 2 + 1 2 I 2 , (1) for both objects. (No slipping implies no frictional losses.) Since v = r (no slipping) oldmidterm 03 KIM, JI Due: Oct 17 2007, noon 2 the final kinetic energy is 1 2 mv 2 + 1 2 I v 2 r 2 = v 2 2 parenleftbigg m + I r 2 parenrightbigg . (2) For many simple geometric shapes, I = C mr 2 , (3) where C is a constant depending on the partic ular shape. In this problem, we have a hoop and a cylinder I hoop = mr 2 C hoop = 1 I cylinder = 1 2 mr 2 C cylinder = 1 2 ....
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 Fall '08
 Turner
 Physics, Energy, Friction, Inertia

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