chap12 - CHAPTER 12 ROTATIONAL MOTION ActivPhysics can help...

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Unformatted text preview: CHAPTER 12 ROTATIONAL MOTION ActivPhysics can help with these problems: Activities 7.7, 7.8, 7.10, 7.127.15 Section 12-1: Angular Velocity and Acceleration Problem 1. Determine the angular speed, in rad/s, of (a) Earth about its axis; (b) the minute hand of a clock; (c) the hour hand of a clock; (d) an egg beater turning at 300 rpm. Solution The angular speed is = / t. (a) E = 1 rev 1 d = 2 / 86 , 400 s = 7 . 27 10 5 s 1 . (b) min = 1 rev / 1 h = 2 / 3600 s = 1 . 75 10 3 s 1 . (c) hr = 1 rev / 12 h = 1 12 min = 1 . 45 10 4 s 1 . (d) = 300 rev / min = 300 2 / 60 s = 31 . 4 s 1 . (Note: Radians are a dimensionless angular measure, i.e., pure numbers; therefore angular speed can be expressed in units of inverse seconds.) Problem 2. What is the linear speed (a) of a point on Earths equator? (b) At your latitude? Solution (a) On the equator, v = E R E = (7 . 27 10 5 s 1 ) (6 . 37 10 6 m) = 463 m / s; see Problem 1(a). (b) At latitude , r = R E cos , so v = E r = (463 m / s)cos . Problem 3. Express each of the following in radians per second: (a) 720 rpm; (b) 50 / h; (c) 1000 rev/s; (d) 1 rev/year (which is the angular speed of Earth in its orbit). Solution (a) (720 rev / min)(2 / rev)(min / 60 s) = 24 s 1 = 75 . 4 s 1 . (b) (50 /h )( / 180 )(h / 3600 s) = 2 . 42 10 4 s 1 . (c) (1000 rev / s)(2 / rev) = 2000 s 1 6 . 28 10 3 s 1 . (d) (1 rev / y) 2 / ( 10 7 s) = 2 10 7 s 1 . (See note in solution to Problem 1. The approximate value for 1 y used in part (d) is often handy for estimates, and is fairly accurate; see Chapter 1, Problem 12.) Problem 4. A 25-cm-diameter circular saw blade spins at 3500 rpm. How fast would you have to push a straight hand saw to have the teeth move through the wood at the same rate as the circular saw teeth? Solution The linear speed of the saw teeth is v = r = ( / 30)(s 1 / rpm)(3500 rpm)(0 . 25 m / 2) = 45 . 8 m / s (over 100 mi/h). Problem 5. A wheel turns through 2.0 revolutions while being accelerated from rest at 18 rpm/s. (a) What is the final angular speed? (b) How long does it take to turn the 2.0 revolutions? Solution For constant angular acceleration, (a) Equation 12-11 gives f = radicalbig 2 + 2 ( f ) = radicalBig 0 + 2(18 rev 60 / min 2 )(2 rev) = 65 . 7 rpm , and (b) Equation 12-10 gives f = 0 + 1 2 t 2 , or t = radicalbig 2(2 rev) / (18 rev / 60 s 2 ) = 3 . 65 s . Problem 6. You switch a food blender from its high to its low setting; the blade speed drops from 3600 rpm to 1800 rpm in 1.4 s. How many revolutions does it make during this time?...
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This note was uploaded on 10/31/2010 for the course PHYSICS 69054 taught by Professor Huang during the Fall '09 term at University of California, Berkeley.

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chap12 - CHAPTER 12 ROTATIONAL MOTION ActivPhysics can help...

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