This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: homework 26 HETTIGER, CHRISTOF 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 . 54 m and a mass of 160 kg. How much work is required to get the sphere rolling with an angular speed of 77 rad / s on a horizontal surface? Assume the sphere starts from rest and rolls without slipping. Correct answer: 193636 J (tolerance 1 %). Explanation: The expression for the rotational energy of a body spinning is given by K rotation = 1 2 I 2 = 1 2 (18 . 6624 kg m 2 ) (77 rad / s) 2 = 55324 . 7 J . For a sphere the moment of inertia I about the center is I = 2 5 mr 2 = 2 5 (160 kg) (0 . 54 m) 2 = 18 . 6624 kg m 2 . The expression for the translational energy of a body is given by K translation = 1 2 mv 2 = 1 2 (160 kg) (41 . 58 m / s) 2 = 138312 J . The work (equal to the kinetic energy) is W = 1 2 mv 2 + 1 2 I 2 = 1 2 (160 kg) (41 . 58 m / s) 2 + 1 2 (18 . 6624 kg m 2 ) (77 rad / s) 2 = 193636 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 = 12 m 2 s 2. bardbl vector L bardbl = 24 kg m 2 s correct 3. bardbl vector L bardbl = 9 N m kg 4. bardbl vector L bardbl = 13 . 5 kg m 2 s 2 5. bardbl vector L bardbl = 18 N m kg 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 . 66 g / cm 3 and mass 0 . 8 kg spins on an axis through its center with a period of 2 s. Given V sphere = 4 3 R 3 , what is its angular momentum? Correct answer: 0 . 000857989 kg m 2 / s (toler ance 1 %). Explanation: The definition of density is M V = M 4 3 R 3 , Therefore R = bracketleftbigg 3 M 4 bracketrightbigg 1 3 = bracketleftbigg 3 (0 . 8 kg) 4 (7660 kg / m 3 ) bracketrightbigg 1 3 = 0 . 029214 m . homework 26 HETTIGER, CHRISTOF Due: Apr 1 2008, 4:00 am 2 Using = 2 T = 2 (2 s) = 3 . 14159 s 1 and I = 2 5 M R 2 = 2 5 (0 . 8 kg)(0 . 029214 m) 2 = 0 . 000273106 kg m 2 , we have L I = 4 M R 2 5 T = 4 (0 . 8 kg)(0 . 029214 m) 2 5 (2 s) = . 000857989 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 2 . 3 m 8 . 9 kg 3 4 g = 9 . 8 m / s 2 Calculate the magnitude of the angular mo mentum of the bob about the supporting point....
View Full
Document
 Spring '10
 Turner
 Physics, Mass, Work

Click to edit the document details