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Homework 11-solutions

# Homework 11-solutions - yindeemark(rry82 Homework 11...

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yindeemark (rry82) – Homework 11 – chelikowsky – (59005) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rod can pivot at one end and is free to rotate without friction about a vertical axis, as shown. A force vector F is applied at the other end, at an angle θ to the rod. L m F θ If vector F were to be applied perpendicular to the rod, at what distance d from the axis of rotation should it be applied in order to produce the same torque vector τ ? 1. d = L cos θ 2. d = L sin θ correct 3. d = L 4. d = 2 L 5. d = L tan θ Explanation: The torque the force generates is τ = F L sin θ . Thus the distance is L sin θ . 002 10.0 points Consider two masses of 3 . 2 kg and 7 . 8 kg connected by a string passing over a pulley having a moment of inertia 8 . 6 g · m 2 about its axis of rotation, as in the figure below. The string does not slip on the pulley, and the system is released from rest. The radius of the pulley is 0 . 24 m. 39 cm 0 . 24 m 8 . 6 g · m 2 ω 7 . 8 kg 3 . 2 kg Find the linear speed of the masses after the 7 . 8 kg mass descends through a distance 39 cm. Assume mechanical energy is con- served during the motion. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 77589 m / s. Explanation: Let : I = 8 . 6 g · m 2 = 0 . 0086 kg · m 2 , R = 0 . 24 m , m 1 = 3 . 2 kg , m 2 = 7 . 8 kg , and h = 39 cm = 0 . 39 m . Consider the free body diagrams 7 . 8 kg 3 . 2 kg T 2 T 1 m 2 g m 1 g a a If we neglect friction in the system, the me- chanical energy is constant and the increase in kinetic energy equals the decrease in po- tential energy. Since K i = 0 (the system is initially at rest), the masses have a common speed and v = R ω , Δ K = K f - K i

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yindeemark (rry82) – Homework 11 – chelikowsky – (59005) 2 = 1 2 m 1 v 2 + 1 2 m 2 v 2 + 1 2 I ω 2 , = 1 2 m 1 v 2 + 1 2 m 2 v 2 + 1 2 I parenleftBig v R parenrightBig 2 = 1 2 parenleftbigg m 1 + m 2 + I R 2 parenrightbigg v 2 . From the figure, we see that m 2 loses potential energy while m 1 gains potential energy; i.e. , Δ U 2 = - m 2 g h and Δ U 1 = m 1 g h . Applying the principle of conservation of energy in the form Δ K + Δ U 1 + Δ U 2 = 0 1 2 parenleftbigg m 1 + m 2 + I R 2 parenrightbigg v 2 = m 2 g h - m 1 g h . v = radicaltp radicalvertex radicalvertex radicalbt 2 ( m 2 - m 1 ) g h m 1 + m 2 + I R 2 . = radicaltp radicalvertex radicalvertex radicalvertex radicalbt 2 (7 . 8 kg - 3 . 2 kg) (9 . 8 m / s 2 ) 3 . 2 kg + 7 . 8 kg + 0 . 0086 kg · m 2 (0 . 24 m) 2 × 0 . 39 m = 1 . 77589 m / s .
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Homework 11-solutions - yindeemark(rry82 Homework 11...

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