Torque &amp; Rolling

# Torque &amp; Rolling - adamkiewicz(ja28724 – Torque...

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Unformatted text preview: adamkiewicz (ja28724) – Torque Rolling – Morozova – (11102) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A uniform rod has length 77 m and mass 8 kg . A mass of 8 kg is attached at one end. The other end of the rod is pivoted about the horizontal axis at O . 8 kg 8 kg 77 m 8 k g 8 kg pivot O 42 ◦ Determine the torque about O immediately after the rod-plus-mass system is released from the horizontal position. The accelera- tion of gravity is 9 . 8 m / s 2 . Correct answer: 9055 . 2 kg · m 2 / s 2 . Explanation: Let : m = m = 8 kg , L = 77 m , and θ = 42 ◦ . There are two forces on the system (the weight of the rod which can be calculated at the center of the rod, and the weight of the mass, so the torque is given by τ = mg parenleftbigg L + 1 2 L parenrightbigg = parenleftbigg 3 2 parenrightbigg mg L = parenleftbigg 3 2 parenrightbigg (8 kg) (9 . 8 m / s 2 ) (77 m) = 9055 . 2 kg · m 2 / s 2 . 002 (part 2 of 3) 10.0 points After the rod-plus-mass system is released, it rotates freely about the point O . Determine its angular velocity ω as the rod passes through the vertical direction. Correct answer: 0 . 53513 s − 1 . Explanation: The moment of inertia of the system about the pivot is given by I = 1 3 mL 2 + mL 2 = 4 3 mL 2 . The center of mass of the system drops a height of 3 4 L as it passes through the vertical direction. By equating the gain of rotational energy and the loss of potential energy, we have 1 2 parenleftbigg 4 3 mL 2 parenrightbigg ω 2 = (2 m ) 3 4 g L ω = radicalbigg 9 g 4 L = radicalBigg 9 (9 . 8 m / s 2 ) 4 (77 m) = . 53513 s − 1 ....
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Torque &amp; Rolling - adamkiewicz(ja28724 – Torque...

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