HW 010 - hanna(brh687 – hk 10 – Opyrchal –(11105 1 This print-out should have 13 questions Multiple-choice questions may continue on the next

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Unformatted text preview: hanna (brh687) – hk 10 – Opyrchal – (11105) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A simple pendulum consists of a small object of mass 4 . 1 kg hanging at the end of a 1 . 8 m long light string that is connected to a pivot point. Find the magnitude of the torque (due to the force of gravity) about this pivot point when the string makes a 3 . 06072 ◦ angle with the vertical. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 3 . 86168 N · m. Explanation: Let : m = 4 . 1 kg , L = 1 . 8 m , and θ = 3 . 06072 ◦ . 1 . 8 m 4 . 1 kg mg 1 . 8 m The torque is τ = F d = mg L sin θ = (4 . 1 kg)(9 . 8 m / s 2 )(1 . 8 m) sin 3 . 06072 ◦ = 3 . 86168 N · m . 002 10.0 points A uniform horizontal rod of mass 1 . 1 kg and length 2 . 2 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I = mℓ 2 12 . 2 . 2 m F 9 9 ◦ pivot 1 . 1 kg If a 2 . 1 N force at an angle of 99 ◦ to the hor- izontal acts on the rod as shown, what is the magnitude of the resulting angular accelera- tion about the pivot point? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 4 . 11056 rad / s 2 . Explanation: Let : ℓ = 2 . 2 m , m = 1 . 1 kg , θ = 99 ◦ , and F = 2 . 1 N . By the parallel axis theorem, the moment of inertia of a stick pivoted at the end is I = I cm + md 2 = 1 12 mℓ 2 + m parenleftbigg ℓ 2 parenrightbigg 2 = 1 3 mℓ 2 . The sum of the torques (counterclockwise ro- tation defined as positive) is summationdisplay τ = F ℓ sin θ − mg ℓ 2 = I α F ℓ sin θ − mg l ℓ 2 = parenleftbigg 1 3 mℓ 2 parenrightbigg α 6 F sin θ − 3 mg = 2 mℓ α vectorα = 6 F sin θ − 3 mg 2 mℓ Since 6 F sin θ − 3 mg = 6 (2 . 1 N) sin99 ◦ − 3 (1 . 1 kg) (9 . 8 m / s 2 ) = − 19 . 8951 N , then bardbl vectorα bardbl = vextendsingle vextendsingle vextendsingle − 19 . 8951 N vextendsingle vextendsingle vextendsingle 2 (1 . 1 kg) (2 . 2 m) = 4 . 11056 rad / s 2 . hanna (brh687) – hk 10 – Opyrchal – (11105) 2 003 10.0 points An Atwood machine is constructed using a...
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This note was uploaded on 04/07/2012 for the course PHYSICS 121 taught by Professor James during the Spring '12 term at University of Texas at Austin.

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HW 010 - hanna(brh687 – hk 10 – Opyrchal –(11105 1 This print-out should have 13 questions Multiple-choice questions may continue on the next

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