# hw26_s - homework 26 – YOO HEE – Due Apr 1 2008 4:00 am...

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Unformatted text preview: homework 26 – YOO, HEE – Due: Apr 1 2008, 4:00 am 1 Question 1, chap 13, sect 2. part 1 of 1 10 points A solid sphere has a radius of 0 . 69 m and a mass of 190 kg. How much work is required to get the sphere rolling with an angular speed of 99 rad / s on a horizontal surface? Assume the sphere starts from rest and rolls without slipping. Correct answer: 620612 J (tolerance ± 1 %). Explanation: The expression for the rotational energy of a body spinning is given by K rotation = 1 2 I ω 2 = 1 2 (36 . 1836 kg m 2 ) (99 rad / s) 2 = 177318 J . For a sphere the moment of inertia I about the center is I = 2 5 mr 2 = 2 5 (190 kg) (0 . 69 m) 2 = 36 . 1836 kg m 2 . The expression for the translational energy of a body is given by K translation = 1 2 mv 2 = 1 2 (190 kg) (68 . 31 m / s) 2 = 443294 J . The work (equal to the kinetic energy) is W = 1 2 mv 2 + 1 2 I ω 2 = 1 2 (190 kg) (68 . 31 m / s) 2 + 1 2 (36 . 1836 kg m 2 ) (99 rad / s) 2 = 620612 J . Question 2, chap 13, sect 4. part 1 of 1 10 points A 2 kg object moves in a circle of radius 4 m at a constant speed of 3 m / s. A net force of 4 . 5 N acts on the object. What is the magnitude of the angular mo- mentum bardbl vector L bardbl of the object with respect to an axis perpendicular to the circle and through its center? 1. bardbl vector L bardbl = 18 N m kg 2. bardbl vector L bardbl = 12 m 2 s 3. bardbl vector L bardbl = 13 . 5 kg m 2 s 2 4. bardbl vector L bardbl = 9 N m kg 5. bardbl vector L bardbl = 24 kg m 2 s correct Explanation: The angular momentum is bardbl vector L bardbl = mv r = (2 kg) (3 m / s)(4 m) = 24 kg · m 2 s . Question 3, chap 13, sect 4. part 1 of 1 10 points A solid steel sphere of density 7 . 51 g / cm 3 and mass 0 . 2 kg spins on an axis through its center with a period of 2 . 5 s. Given V sphere = 4 3 π R 3 , what is its angular momentum? Correct answer: 6 . 90024 × 10 − 5 kg m 2 / s (tol- erance ± 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 . 2 kg) 4 π (7510 kg / m 3 ) bracketrightbigg 1 3 = 0 . 0185254 m . homework 26 – YOO, HEE – Due: Apr 1 2008, 4:00 am 2 Using ω = 2 π T = 2 π (2 . 5 s) = 2 . 51327 s − 1 and I = 2 5 M R 2 = 2 5 (0 . 2 kg)(0 . 0185254 m) 2 = 2 . 74552 × 10 − 5 kg m 2 , we have L ≡ I ω = 4 π M R 2 5 T = 4 π (0 . 2 kg)(0 . 0185254 m) 2 5 (2 . 5 s) = 6 . 90024 × 10 − 5 kg m 2 / s . Question 4, chap 13, sect 4. part 1 of 1 10 points A small metallic bob is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. v 4 m 8 . 2 kg 2 7 ◦ g = 9 . 8 m / s 2 Calculate the magnitude of the angular mo- mentum of the bob about the supporting point....
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## This note was uploaded on 05/04/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw26_s - homework 26 – YOO HEE – Due Apr 1 2008 4:00 am...

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