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Kaplunovsky - practice 11

# Kaplunovsky - practice 11 - practice 11 Due noon Question 1...

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practice 11 – – Due: Apr 10 2007, noon 1 Question 1, chap 13, sect 2. part 1 of 1 10 points A constant horizontal force of 120 N is ap- plied to a lawn roller in the form of a uniform solid cylinder of radius 0 . 35 m and mass 13 kg. The acceleration of gravity is 9 . 8 m / s 2 . M R F If the roller rolls without slipping, calculate the acceleration of the center of mass. ( I cm = 1 2 M R 2 ) Correct answer: 6 . 15385 m / s 2 (tolerance ± 1 %). Explanation: Basic Concepts: summationdisplay vector F = mvector a summationdisplay vector τ = Ivectorα If we choose the center of mass as the axis, summationdisplay F = F f = M a τ = fR = I cm α = 1 2 M R 2 a R then you can solve these two equations to get a. The another way to do this is we can choose the contact point as the axis, using I = 1 2 M R 2 + M R 2 = 3 2 M R 2 , we find 3 2 M R 2 a R = F R . Therefore a = 2 F 3 M (1) = 2 (120 N) 3 (13 kg) = 6 . 15385 m / s 2 . Question 2, chap 13, sect 2. part 1 of 3 7 points Consider a thick cylindrical shell rolling down along an incline. It has a mass m , an outer radius R and a moment of iner- tia I cm = 1 2 mR 2 about its center of mass. The incline is at an angle θ to the hori- zontal. Its length is s and its height is h . s h θ v A A The equation of motion τ = as the shell is rolling down the incline is given by (using the contact point as the pivot point) 1. m g R = I cm α 2. m g R tan θ = ( I cm + mR 2 ) α 3. m g R cot θ = ( I cm + mR 2 ) α 4. m g R sin θ = ( I cm + mR 2 ) α correct 5. m g R cos θ = ( I cm + mR 2 ) α 6. m g R sin θ = I cm α 7. m g R cos θ = I cm α 8. m g R tan θ = I cm α 9. m g R cot θ = I cm α 10. m g R = ( I cm + mR 2 ) α Explanation: Basic Concepts: Energy conservation U i + K i = U f + K f (1) where the KE terms include both rotational and translational kinetic energy. Equation of motion τ =

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practice 11 – – Due: Apr 10 2007, noon 2 θ θ mg sin θ mg P The torque about the contact point P is given by τ = R m g sin θ = I p α . Using the law of parallel axes, I p = I cm + m R 2 . This leads to the choice of m g R sin θ = ( I cm + m R 2 ) α. Question 3, chap 13, sect 2. part 2 of 3 7 points The linear acceleration a of the shell as it rolls down the incline is given by 1. a = 9 5 g sin θ 2. a = 5 9 g sin θ 3. a = 2 3 g cos θ 4. a = 5 9 g cos θ 5. a = 9 5 g cos θ 6. a = g cos θ 7. a = 3 2 g sin θ 8. a = 3 2 g cos θ 9. a = 2 3 g sin θ correct 10. a = g sin θ Explanation: Using the law of parallel axes, one finds that the moment of inertia of the shell about P I = I cm + mR 2 = parenleftbigg 1 2 parenrightbigg m R 2 + m R 2 = parenleftbigg 3 2 parenrightbigg m R 2 Now we apply the equation of motion τ = I α I α = parenleftbigg 3 2 parenrightbigg m R 2 α = parenleftbigg 3 2 parenrightbigg m R a , and τ = R m g sin θ . This leads to a = parenleftbigg 2 3 parenrightbigg g sin θ . Question 4, chap 13, sect 2. part 3 of 3 7 points Release the shell from rest at the top of the incline.
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