Department of Natural Sciences Clayton State University. Physics 1111 Quiz 8. a. Find the average angular acceleration of the windmill.

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1 November 5, 2007 Physics 1111 Quiz 8 Name SOLUTION 1. Express 27.5 o angle in radians. (27.5 o ) /(180 o ) = rad 2. As the wind dies, a windmill that was rotating at 3.60 rad/s comes to a full stop in 5.00 s. t a. Find the average angular acceleration of the windmill. ( t = (0 (3.60 rad/s))/(5.00 s) = rad/s 2 b. Through what angle did the windmill rotate before coming to the full stop? ) ) = ( ) ) = ( rad/s rad/s 2 ) = 9.00 rad c. How many rotations does it correspond to? (9.00 rad)(1 rev)/(2 rad) = 1.43 rev Clayton College & State University Physics 1111 Quiz 7 March 23, 2005

2 Name SOLUTION A cooling fan is turned off when it is running at 850 rev/min. It turns 1500 revolutions before it comes to a stop. a. What is the initial angular velocity of the fan in rad/s? = (850 rev/min) x (2 rad)/(1 rev) x (1 min)/(60.0 s) = 89.0 rad/s f = 0 rad/s i = 89.0 rad/s b. What was the fan s angular acceleration, assumed constant? f - i = `(1500 rev) x (2 rad/rev) = 3000 rad f 2 = i ( f - i ) f 2 - i 2 )/ [2 ( f - i)] = = rad/s 2 f = i + t f - i ) / = t t = 212 s c. How long did it take the fan to come to a complete stop? Physics 1111 Quiz 8 April 5, 2006 Name SOLUTION 1. A flywheel with a radius of m starts from rest and accelerates with a constant angular acceleration of rad/s 2. After the flywheel has turned through 60.0 o,

3 a. What is the magnitude of the centripetal acceleration of the point on the rim of the flywheel? ) rad/s 2 ) 1.12 rad/s a cp = r = (1.12 rad/s) (0.300 m) = m/s 2 b. What is the magnitude of the tangential acceleration of the same point? a t = r = (0.600 rad/s 2 ) (0.300 m) = m/s 2 c. What is the magnitude of the total acceleration of the point? a tot = (a cp 2 + a t 2 ) 1/2 = m/s 2 Physics 1111 Quiz 8 April 2, 2007 Name SOLUTION The blades of a fan running at low speed turn at 250 rpm. When the fan is switched to high speed, the rotation rate increases uniformly to 350 rpm in 5.75 s. a. Express initial and final angular velocities of the fan in rad/s. o = (250 rpm)(2 rad/1 rev)(1 m/60 s) = 26.2 rad/s = (350 rpm)(2 rad/1 rev)(1 m/60 s) = 36.7 rad/s b. What is the magnitude of the angular acceleration of the blades? t t

4 t (36.7 rad/s 26.2 rad/s)/(5.75 s) = 1.83 rad/s 2 c. How many revolutions do the blades go through while the fan is accelerating? ) ) = ( ) = ((36.7 rad/s) 2 - (26.2 rad/s) 2 )/(2 (1.83 rad/s 2 )) = 180 rad = 28.7 rev Physics 1111 Quiz 11 April 16, 2008 Name SOLUTION You are located on the outer rim of a merry-go-round of diameter 20.0 m and are orbiting at 6.00 rev/min. a. What is your angular speed in rad/s? = (6.00 rev/min)(2 rad/rev)(1 min/60 sec) = rad/s b. What is your tangential speed? V t = r = (0.628 rad/s)(10.0 m) = 6.28 m/s c. What is the magnitude of your centripetal acceleration? a cp = V t 2 / r = (6.28 m/s) 2 /(10.0 m) = 3.95 m/s 2 d. Assuming that your mass is 60.0 kg, how much of the centripetal force is necessary to keep you on merrry-go-round? F cp = m a cp = (60.0 kg)(3.95 rad/s 2 ) = 237 N

5 Clayton College & State University Physics 1111 Quiz 10 July 1, 2004 Name SOLUTION You are located at the outer rim of a merry-go-round of diameter 20.0 m and are orbiting at 6.00 rev/min. a. What is your angular speed in rad/s? = (6.00 rev/m) x (2 rad)/(1 rev) x (1 min)/(60.0 s) = rad/s b. What is your tangential speed in m/s? V t = r = (10.0 m) (0.628 rad/s) = 6.28 m/s The merry-go-round starts to slow down and comes to a complete stop after making 10 full revolutions. What is the angular acceleration of the merry-go round while it slows down? Assume that the angular acceleration is constant. f = 0 rad/s i = rad/s f - i = `(10 rev) x (2 rad/rev) = 20 rad f 2 = i ( f - i ) f 2 - i 2 )/ 2 ( f - i) = = rad/s 2 Physics 1111 Quiz 9 July 20, 2005 Name SOLUTION

6 A 70-cm-diameter wheel accelerates uniformly about its center from 130 rpm to 280 rpm in 4.0 s. Determine (a) its angular acceleration, and o = (130 rpm)(2 rad / 1 rev)(1min /60 s ) = 13.6 rad/s = (280 rpm)(2 rad / 1 rev)(1min /60 s ) = 29.3 rad/s = o )/t = (29.3 rad/s 13.6 rad/s) / (4.00 s) = 3.93 rad/s 2 (b) the radial and tangential components of the linear acceleration of a point on the edge of the wheel 2.0 s after it has started accelerating. a t = R = (3.93 rad/s 2 )(0.350 m) = 1.37 m/s 2 o + t = (0 rad/s) + (3.93 rad/s 2 )(2.00 s) = 7.86 rad/s a c = R 2 = (7.86 rad/s)(0.350 m) 2 = m/s 2 Physics 1111 Quiz 9 July 11, 2007 Name SOLUTION 1. Express 47.5 o angle in radians. (47.5 o x rad)/180 o = rad

7 2. As the wind dies, a windmill that was rotating at 2.50 rad/s comes to a full stop in 6.00 s. Find the average angular acceleration of the windmill. av = t = ( rad/s)/(6.00 s) = rad/s 2 3. A disk is rotating about its center with a constant angular velocity of 10.3 rad/s. Through what angular displacement does a point on the rim of the disk go during 15.0 s time interval? av = t av t = (10.3 rad/s)(15.0 s) = 155 rad

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