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HW04-solutions - Le(tl8426 HW04 gentle(56245 This print-out...

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Le (tl8426) – HW04 – gentle – (56245) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A point on the outer rim of a tire on a moving vehicle exhibits uniform circular motion with a frequency of 28 Hz. The diameter of the tire is 39 . 2 cm. How fast is the car moving? Correct answer: 124 . 136 km / hr. Explanation: You first need to find the length λ of one cycle, which is the circumference of the tire. Thus λ = πd Thus the velocity is defined by v = Conversion for v: 1 s cm 1 1 m 100 cm 1 km 1000 m 3600 s 1 hr = km / hr 002(part1of2)10.0points Young David, who slew Goliath, experi- mented with slings before tackling the gi- ant. He found that with a sling of length 0 . 724 m, he could revolve the sling at the rate of 10 . 9 rev / s. If he increased the length to 0 . 963 m, he could revolve the sling only 6 . 9 rev / s. a) What is the larger of the two linear speeds? Correct answer: 49 . 5844 m / s. Explanation: BasicConcept: The angular velocity ω is defined as ω v r , in units of rad s . Solution: v = r ω, so v 1 = r 1 ω 1 = (0 . 724 m) (10 . 9 rev / s) parenleftbigg 2 π rad rev parenrightbigg = 49 . 5844 m / s . and v 2 = r 2 ω 2 = (0 . 963 m) (6 . 9 rev / s) parenleftbigg 2 π rad rev parenrightbigg = 41 . 7499 m / s . Therefore the greater linear speed is 49 . 5844 m / s. 003(part2of2)10.0points b) Using the sling length 0 . 963 m, what is the centripetal acceleration at 6 . 9 rev / s? Correct answer: 1810 . 02 m / s 2 . Explanation: The centripetal acceleration is given by a 2 = v 2 2 r 2 = r 2 ω 2 2 = (0 . 963 m) (6 . 9 rev / s) 2 parenleftbigg 2 π rad 1 rev parenrightbigg 2 = 1810 . 02 m / s 2 . 004(part1of2)10.0points The orbit of a Moon about its planet is ap- proximately circular, with a mean radius of 3 . 89 × 10 8 m. It takes 13 . 7 days for the Moon to complete one revolution about the planet. Find the mean orbital speed of the Moon. Correct answer: 2064 . 88 m / s. Explanation: Dividing the length C = 2 πr of the trajectory of the Moon by the time T = 13 . 7 days = 1 . 18368 × 10 6 s
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Le (tl8426) – HW04 – gentle – (56245) 2 of one revolution (in seconds!), we obtain that the mean orbital speed of the Moon is v = C T = 2 π r T = 2 π (3 . 89 × 10 8 m ) 1 . 18368 × 10 6 s = 2064 . 88 m / s . 005(part2of2)10.0points Find the Moon’s centripetal acceleration. Correct answer: 0 . 0109608 m / s 2 . Explanation: Since the magnitude of the velocity is con- stant, the tangential acceleration of the Moon is zero. The centripetal acceleration is a c = v 2 r = (2064 . 88 m / s ) 2 3 . 89 × 10 8 m = 0 . 0109608 m / s 2 . 006 10.0points A ball on the end of a string is whirled around in a horizontal circle of radius 0 . 308 m. The plane of the circle is 1 . 77 m above the ground. The string breaks and the ball lands 2 . 61 m away from the point on the ground directly beneath the ball’s location when the string breaks. The acceleration of gravity is 9 . 8 m / s 2 . Find the centripetal acceleration of the ball during its circular motion.
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