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PHY 303K - Homework 19

# PHY 303K - Homework 19 - homework 19 FIERRO JEFFREY Due Apr...

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homework 19 – FIERRO, JEFFREY – Due: Apr 6 2008, 11:00 pm 1 Question 1, chap 13, sect 4. part 1 of 1 10 points A solid steel sphere of density 7 . 85 g / cm 3 and mass 0 . 7 kg spins on an axis through its center with a period of 2 . 4 s. Given V sphere = 4 3 π R 3 , what is its angular momentum? Correct answer: 0 . 000563058 kg m 2 / s (toler- ance ± 1 %). Explanation: The definition of density is ρ M V = M 4 3 π R 3 , Therefore R = bracketleftbigg 3 M 4 π ρ bracketrightbigg 1 3 = bracketleftbigg 3 (0 . 7 kg) 4 π (7850 kg / m 3 ) bracketrightbigg 1 3 = 0 . 0277149 m . Using ω = 2 π T = 2 π (2 . 4 s) = 2 . 61799 s 1 and I = 2 5 M R 2 = 2 5 (0 . 7 kg)(0 . 0277149 m) 2 = 0 . 000215072 kg m 2 , we have L I ω = 4 π M R 2 5 T = 4 π (0 . 7 kg)(0 . 0277149 m) 2 5 (2 . 4 s) = 0 . 000563058 kg m 2 / s . Question 2, chap 13, sect 4. part 1 of 1 10 points A force F x = (2 N)ˆ ı + (3 . 5 N)ˆ is applied to an object that is pivoted about a fixed axis aligned along the z coordinate axis. The force is applied at the point x = (4 . 1 m)ˆ ı +(3 . 9 m)ˆ . Find the z -component of the net torque. Correct answer: 6 . 55 N m (tolerance ± 1 %). Explanation: Basic Concept: vector τ = vectorr × vector F Solution: From the definition of torque vector τ = vectorr × vector F = [ x ˆ ı + y ˆ ] × [ F x ˆ ı + F y ˆ ] = [(4 . 1 m)ˆ ı + (3 . 9 m)ˆ ] × [(2 N)ˆ ı + (3 . 5 N)ˆ ] = [(4 . 1 m) (3 . 5 N) (3 . 9 m) (2 N)] ˆ k = (6 . 55 N m) ˆ k In this case, the net torque only has a z - component. Question 3, chap 13, sect 4. part 1 of 2 10 points A particle is located at the vector position vectorr = (1 . 8 m)ˆ ı + (3 . 9 m)ˆ and the force acting on it is vector F = (5 . 4 N)ˆ ı + (1 . 6 N)ˆ  . What is the magnitude of the torque about the origin? Correct answer: 18 . 18 N m (tolerance ± 1 %). Explanation: Basic Concept: vector τ = vectorr × vector F Solution: Since neither position of the par- ticle, nor the force acting on the particle have

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homework 19 – FIERRO, JEFFREY – Due: Apr 6 2008, 11:00 pm 2 the z -components, the torque acting on the particle has only z -component: vector τ = [ x F y y F x ] ˆ k = [(1 . 8 m) (1 . 6 N) (3 . 9 m) (5 . 4 N)] ˆ k = [ 18 . 18 N m] ˆ k . Question 4, chap 13, sect 4. part 2 of 2 10 points What is the magnitude of the torque about the point having coordinates [ a, b ] = [(4 m) , (6 m)]? Correct answer: 7 . 82 N m (tolerance ± 1 %). Explanation: Reasoning similarly as we did in the pre- vious section, but with the difference that relative to the point [(4 m) , (6 m)] the y - component of the particle is now [ y (6 m)], we have vector τ = { [ x a ] F y [ y b ] F x } ˆ k = { [(1 . 8 m) (4 m)] [1 . 6 N] [(3 . 9 m) (6 m)] [5 . 4 N] } ˆ k = { 7 . 82 N m } ˆ k .
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PHY 303K - Homework 19 - homework 19 FIERRO JEFFREY Due Apr...

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