Department of Natural Sciences
Clayton State University
November 5, 2007
Physics 1111 – Quiz 8
Name ____SOLUTION_________________________________
1. Express 27.5o angle in radians.
(27.5o)  /(180o) = 0.480 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.
a. Find the average angular acceleration of the windmill.
t
(t = (0 – (3.60 rad/s))/(5.00 s) = - 0.720 rad/s2
b. Through what angle did the windmill rotate before coming to the full stop?
)
) = ()
) = (03.60 rad/s0.720 rad/s2) = 9.00 rad
c. How many rotations does it correspond to?
(9.00 rad)(1 rev)/(2 rad) = 1.43 rev
Department of Natural Sciences
Clayton College & State University
March 23, 2005
Physics 1111 – Quiz 7
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
b. What was the fan’s angular acceleration, assumed constant?
f = 0 rad/s

i = 89.0 rad/s
f -  i = `(1500 rev) x (2 rad/rev) = 3000 rad
f 2 = i 2 + 2  (f - i)
f 2 - i 2)/ [2 (f -  i)] =  = - 0.420 rad/s2
c. How long did it take the fan to come to a complete stop?
f = i +  t
f - i) / = t
t = 212 s
Department of Natural Sciences
Clayton State University
April 5, 2006
Physics 1111 – Quiz 8
Name ____SOLUTION_________________________________
1. A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant
angular acceleration of 0.600 rad/s2. After the flywheel has turned through 60.0o,
a. What is the magnitude of the centripetal acceleration of the point on the rim of
the flywheel?
)
0.600 rad/s2)
1.12 rad/s
acp = r = (1.12 rad/s) (0.300 m) = 0.376 m/s2
b. What is the magnitude of the tangential acceleration of the same point?
at = r = (0.600 rad/s2) (0.300 m) = 0.180 m/s2
c. What is the magnitude of the total acceleration of the point?
atot = (acp2 + at2)1/2 = 0.417 m/s2
Department of Natural Sciences
Clayton State University
April 2, 2007
Physics 1111 – Quiz 8
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
t
(36.7 rad/s – 26.2 rad/s)/(5.75 s) = 1.83 rad/s2
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/s2)) = 180 rad = 28.7 rev
Department of Natural Sciences
Clayton State University
April 16, 2008
Physics 1111 – Quiz 11
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) = 0.628 rad/s
b. What is your tangential speed?
Vt =  r = (0.628 rad/s)(10.0 m) = 6.28 m/s
c. What is the magnitude of your centripetal acceleration?
acp = Vt2 / r = (6.28 m/s)2/(10.0 m) = 3.95 m/s2
d. Assuming that your mass is 60.0 kg, how much of the centripetal force is
necessary to keep you on merrry-go-round?
Fcp = m acp = (60.0 kg)(3.95 rad/s2) = 237 N
Department of Natural Sciences
Clayton College & State University
July 1, 2004
Physics 1111 – Quiz 10
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) = 0.628 rad/s
b. What is your tangential speed in m/s?
Vt = 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 = 0.628 rad/s
f -  i = `(10 rev) x (2 rad/rev) = 20 rad
f 2 = i 2 + 2  (f - i)
f 2 - i 2)/ 2 (f -  i) =  = - 0.00314 rad/s2
Department of Natural Sciences
Clayton State University
July 20, 2005
Physics 1111 – Quiz 9
Name __SOLUTION___________________________________
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/s2
(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.
at =  R = (3.93 rad/s2)(0.350 m) = 1.37 m/s2
o +  t = (0 rad/s) + (3.93 rad/s2)(2.00 s) = 7.86 rad/s
ac =  R2 = (7.86 rad/s)(0.350 m)2 = 0.963 m/s2
Department of Natural Sciences
Clayton State University
July 11, 2007
Physics 1111 – Quiz 9
Name ___SOLUTION__________________________________
1. Express 47.5o angle in radians.
(47.5o x  rad)/180o = 0.829 rad
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 = (0 – 2.50 rad/s)/(6.00 s) = -0.417 rad/s2
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