HW9 - boland (kbb544) – HW9 – mackie – (10611) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: boland (kbb544) – HW9 – mackie – (10611) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A can of soup has a mass 120 g, height 17 cm, and diameter 2 . 8 cm. It is placed at rest on the top of an incline that is 1 . 4 m long at 28 ◦ to the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . Calculate the moment of inertia of the can if it takes 0 . 88 s to reach the bottom of the incline. Correct answer: 6 . 40817 × 10 − 6 kg m 2 . Explanation: The speed of the center of mass of the can at the bottom of the incline is v f = 2¯ v = 2 L t . Thus the angular speed at the bottom of the incline is ω f = v f R . Using energy conservation, we have m g L sin θ = 1 2 m v 2 f + 1 2 I ω 2 f . Solving for I , I = 2 m ω 2 f bracketleftbigg g L sin θ- 1 2 v 2 f bracketrightbigg = 2 (120 g) (227 . 273 rad / s) 2 × bracketleftbigg (9 . 8 m / s 2 )(1 . 4 m) sin(28 ◦ )- 1 2 (3 . 18182 m / s) 2 bracketrightbigg = 6 . 40817 × 10 − 6 kg m 2 . 002 10.0 points A solid steel sphere of density 7 . 5 g / cm 3 and mass 0 . 6 kg spin on an axis through its center with a period of 3 . 5 s. Given V sphere = 4 3 π R 3 , what is its angular momentum? Correct answer: 0 . 000307839 kg m 2 / s. 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 . 6 kg) 4 π (7500 kg / m 3 ) bracketrightbigg 1 / 3 = 0 . 0267301 m . Using ω = 2 π T = 2 π (3 . 5 s) = 1 . 7952 s − 1 and I = 2 5 M R 2 = 2 5 (0 . 6 kg)(0 . 0267301 m) 2 = 0 . 000171479 kg m 2 , we have L ≡ I ω = 4 π M R 2 5 T = 4 π (0 . 6 kg)(0 . 0267301 m) 2 5 (3 . 5 s) = 0 . 000307839 kg m 2 / s . 003 10.0 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. boland (kbb544) – HW9 – mackie – (10611) 2 v 9 . 8 m / s 2 4 m 9 kg 1 9 ◦ Calculate the magnitude of the angular mo- mentum of the bob about a vertical axis through the supporting point. The acceler- ation of gravity is 9 . 8 m / s 2 . Correct answer: 24 . 5694 kg · m 2 / s. Explanation: Let : ℓ = 4 m , θ = 19 ◦ , g = 9 . 8 m / s 2 , and m = 9 kg ....
View Full Document

This note was uploaded on 02/20/2011 for the course PHYS 1062 taught by Professor Mackie during the Spring '11 term at Temple.

Page1 / 6

HW9 - boland (kbb544) – HW9 – mackie – (10611) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online