ukw1ymnEAaWV_1210016133_jwk572

ukw1ymnEAaWV_1210016133_jwk572 - oldmidterm 03 – KIM JI...

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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 : τ = 28 . 3 N · m , I = 0 . 11 kg · m 2 , and θ = 16 . 4 rev . The work done on the grindstone is W net = vector F vectors = F · ( r θ ) = ( F · r ) θ = τ · θ Thus, W net = 1 2 I w 2 f − 1 2 I w 2 i so τ θ = 1 2 I ω 2 − and ω = radicalbigg 2 θ τ I = radicalBigg 2 (16 . 4 rev)(28 . 3 N · m) . 11 kg · m 2 2 π rev · rev 2 π = 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 ω 2 , (1) for both objects. (No slipping implies no frictional losses.) Since v = r ω (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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ukw1ymnEAaWV_1210016133_jwk572 - oldmidterm 03 – KIM JI...

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