HW #10-solutions

# HW #10-solutions - wolz(cmw2833 – HW#10 – Antoniewicz...

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Unformatted text preview: wolz (cmw2833) – HW #10 – Antoniewicz – (57420) 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 Consider the problem of the solid sphere rolling down an incline without slipping. The incline has an angle θ , the sphere’s length up the incline is ℓ , and its height is h . At the beginning, the sphere of mass M and radius R rests on the very top of the incline. M μ ℓ θ h What is the minimum coefficient of friction such that the sphere rolls without slipping? The moment of inertia of a sphere with respect to an axis through its center is 2 5 M R 2 . 1. μ = 5 7 cos θ 2. μ = 3 7 tan θ 3. μ = 2 7 sin θ 4. μ = 3 7 sin θ 5. μ = 5 7 tan θ 6. μ = 2 7 tan θ correct 7. μ = 3 5 cos θ 8. μ = 2 7 cos θ Explanation: Consider the forces acting on the sphere: mg cos θ N f mg sin θ Using the parallel-axis theorem I = 2 5 M R 2 + M R 2 = 7 5 M R 2 , and because the sphere rolls without slipping, α = a R . With the origin at the point of contact between the sphere and the incline surface, summationdisplay τ : M g R sin θ = I α M g R sin θ = 7 5 M R 2 a R a = 5 7 g sin θ . The net force along the direction of the incline is summationdisplay F = M g sin θ- f = M parenleftbigg 5 7 g sin θ parenrightbigg , where f ≤ μ N = μ M g cos θ is the minimum “no slipping” criterion. Then M g sin θ- μ M g cos θ = M parenleftbigg 5 7 g sin θ parenrightbigg μ cos θ = parenleftbigg 1- 5 7 parenrightbigg sin θ μ = 2 7 tan θ . 002 10.0 points A solid cylinder of mass M = 14 kg, radius R = 0 . 42 m and uniform density is pivoted on a frictionless axle coaxial with its symmetry axis. A particle of mass m = 3 . 2 kg and initial velocity v = 14 m / s (perpendicular to the cylinder’s axis) flies too close to the wolz (cmw2833) – HW #10 – Antoniewicz – (57420) 2 cylinder’s edge, collides with the cylinder and sticks to it. Before the collision, the cylinder was not ro- tating. What is the magnitude of its angular velocity after the collision? Correct answer: 10 . 4575 rad / s. Explanation: Basic Concept: Conservation of Angu- lar Momentum, L particle z + L cylinder z = const . The axle allows the cylinder to rotate without friction around a fixed axis but it keeps this axis fixed. Let the z coordinate axis run along this axis of rotation; then the axle may exert arbitrary torques in x and y directions but τ z ≡ 0. Consequently, the z componenent of the angular momentum must be conserved, L z = const, hence when the particle collides with the cylinder L before z, part + L before z, cyl = L z, net = L after z, part + L after z, cyl ....
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## This note was uploaded on 05/02/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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HW #10-solutions - wolz(cmw2833 – HW#10 – Antoniewicz...

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