Physics Honors – Dr. Naumoff – Final Review, Torque and Static Equilibrium
Torque=τ = Force applied x lever arm=F x r
Torque – force not perpendicular to lever arm τ = F x r x sinθ
Conditions for equilibrium
The sum of all the forces (net force) acting on an object is zero.  Fi  0
i
The sum of all the torques (net torque) acting on an object is zero.  i  0
i
Total Clockwise Torques = Total Counterclockwise Torques (about a chosen axis).
1. A force of magnitude F =20.0 N is applied at a distance = 2.0m from the axis of rotation
in an orientation where r makes the angle θ =90o with respect to the line of action of the
force. What is magnitude of the torque at the axis of rotation?
  Fr  20.0  2.0  40.Nm
2. A force of 15.0 N is applied to a 50.0cm wrench at an angle of 45o to the lever arm. What
is magnitude of the torque at the axis of rotation?
  Frsin  15.0  0.50  sin45o  5.3Nm
3. m1=1.0kg L1=1.0m; m2=1.0kg L2=0.5m; m3=3.0kg L3=?; mass of board =5.0kg. The
fulcrum is supporting the board at its center.
a. What is L3?
 cc  9.81 11  1 0.5   14.7 Nm   c
 m3 g  L3  L3 
c
m3 g

14.7
 0.50m
3.0  9.81
b. What is the total force exerted upward by the fulcrum?
Fup  Fdown  Fup   9.811  1  3  5  98.1N
4. Use the picture above for the following problems.
m1=1.0kg; L=1.50m; mass of board =5.0kg ; x=.50m
What is F1?
Choose the left end of the board as the axis of rotation.
 cc   c
 c  9.81 1.0  0.50    5.0  0.75    41.69 Nm   cc
 cc
41.69
 27.8 N
1.5
1.5
F1  Fdown  F2   9.811.0  5.0   27.8  31.1N
 cc  1.5 F2   F2 

5. A board is supported at each end. The board is 3.0m long and has a weight of 20.0N.
There is a weight of 40.0N at a distance of 1.0m from the left end of the board. A weight
of 60.0N is located at a distance of 2.00 m from the left end of the board. What are the
upward forces at the ends of the board?
Choose the left end of the board as the axis of rotation.
 cc   c
 c   40.0 1.0    20.0 1.5    60.0  2   190 Nm   cc
Fup  Fdown
 cc
190
 63.3N
L 3.0
 F1  F2  40  20  60  120 N
 cc  F2 L  F2 

F1  Fdown  F2  120  63.3  56.7 N
Dr. Naumoff Physics Honors – Final review
Rotational kinematics and dynamics.
1 revolution  360o  2 radians
Angular Displacement  
f

  f  i

t
t f  ti

  f  i

t
t f  ti
For a rolling object
vtranslational  r
 =I
  net torque
i
  o   t
For an object rotating in place
K=
  o t 
 2  o2  2
1
2
  (o   )t
For a rolling object
1 2
I
2
K=
I  moment of inertia
1 2
t
2
1 2 1 2
1 2
1 v
mv  I  =
mv 
I 
2
2
2
2 r
2
  angular acceleration
the angular momentum of a rotating object is represented by the letter L
kg m2
L=I unit;
I  moment of inertia
  angular velocity
s
angular momentum is conserved L=L
W
W=τ  Δθ 
unit; N m=J power(P) P=
W=Pt
t
1) A child is sitting on the outer edge of a merry-go-round that is 18 m in diameter. If the merrygo-round makes 4.9 rev/min, what is the velocity of the child in m/s?
m
 18m  4.9rev  2 rad  1min 
v  r  



  4.62
s
 2  1min  1rev  60s 
2) Through what angle in degrees does a 33 rpm record turn in 1.0 s?
o
 33rot   360   1min 
o
  t  


 1s   198
 1min   1rot   60s 
3) A wheel accelerates from rest to 10 rad/s at a rate of 15 rad/s2. Through what angle (in
radians) did the wheel turn while accelerating?
2
2
 rad   rad 
2
2
10
 0




s   s 

2
2
o
  o  2   

 3.33rad
2
 rad 
 2  15 2 
s 

4) A wheel is accelerating from rest at a constant angular acceleration of   1.0
rad
.
s2
a. What is ϴ at t=1.0s?
1
1
2
 t 2  0  11  0.50rad
2
2
b. What is ω at 1.0s?
rad
  o   t  0  11  1
s
c. If the wheel has a moment of inertia I  2.0kg m2 .What is the torque acting on the
  ot 
wheel?
  I   2.01.0  2.0 Nm
d. If the wheel has a diameter of 2.0 m, what is the tangential velocity of a point on its rim
at t=3.0s?
  o   t  0  1 3  3
rad
s
m
2
vtangential  r     3  3
s
2
4) A hoop with a mass of 2.0 kg, and a radius of 0 .10m is rotating about an axis through its center
at 10.0
rad
2
. The moment of inertia for a hoop is I  mr .
s
a. Calculate the hoop’s moment of inertia.
I  mr 2   2.0  0.10   0.02kg m 2
2
b. Calculate the hoop’s kinetic energy.
K
1 2 1
2
I    0.02 10.0   1.0 J
2
2
5) A hoop with a mass of 10.0 kg, and a radius of 0 .10m is rotating about the axis through its center
at 15.0
rad
2
. The moment of inertia for a hoop is I  mr .
s
a. Calculate the hoop’s moment of inertia.
I  mr 2  10.0  0.10   0.10kg m 2
2
b. Calculate the hoop’s kinetic energy.
K
1 2 1
2
I    0.10 15.0   11.25 J
2
2
c. If the hoop were rolling on the ground at the above angular velocity, what would be its
translational velocity?
vtangential  vtranslational  r   0.10 15   1.5
m
s
d. If the hoop were rolling on the ground at the above angular velocity, what would be its
kinetic energy?
2
1
1
1 2
1 v
K= mv 2  I  2 =
mv 
I 
2
2
2
2 r
1
1
2
2
K= 10 1.5    0.10 15   22.5 J
2
2
1. A wave has a period of 2.50 s and a wavelength of 3.00 m.
a. What is its speed?
v  f    0.40  3.0   1.2
m
s
b. What is its frequency?
f 
1
1

 0.40 Hz
T 2.5
c. The above wave transits to a different medium. Its frequency is unchanged,
but its wavelength becomes 4.0 m. What is its new speed?
v  f    0.40  4.0   1.6
m
s
2. Waves on a vibrating string are moving at 10.0
m
. The frequency of vibration is
s
15 Hz. The tension in the string is 100.0N. What is the string’s
v
F


mass
ratio  ?
length
mass
F
F
100
kg
 v2     2 
 1.0
2
length

v
m
10 
3. A closed tube is 2.50 m long. Assume that the speed of sound is 343m/s.
a. What is the lowest frequency at which the column of air in this pipe will vibrate?
1  4 L   4  2.5   10m
f1 
v

343
 34.3Hz
10
1
b. What is the next longest closed pipe that will vibrate at this frequency? 7.5m
c. What is the minimal length of an open pipe that will vibrate at this frequency?
5.0m
4. (10 points) A steel string is 1.50m long and undergoing a tension of 500.0N. The
value of µ for the string is 1.5x10-2
kg
.
m
a. What is the value of the fundamental frequency of this string?
F
v

=
500.0
m
 182.6
2
1.5 10
s
1  2 L   2 1.5   3.0m
f1 
v
1

182.6
 60.9 Hz
3.0
b. What are the values of the next three allowable harmonics?
121.8Hz; 182.7Hz;243.6Hz
5. Sketch the first three allowable harmonics for a 1.0m closed pipe. Label lengths,
nodes, and antinodes.
6. In a different universe, light in a vacuum travels at a different speed than in our universe.
Scientists have set up a Michelson experiment to measure the speed of light using an 10
sided rotating mirror and two fixed mirrors 30 000 meters from the rotating mirror. A
clear image is obtained when the mirror is rotating at 200.0 rps (rotations per second).
a. What is the speed of light in that universe?
Let f  the number of rotations per second or frequency
1
1
then the time required for one rotation is  
 5.0 10-3 s
f 200
Time required for the mirror to advance one tenth of a rotation
1
5.0 10-3 s
t

 5.0 10-4 s
rps  number of faces
10
t is also the time required for light to make a round trip between the mirrors
2d  2  30 000 
m

 1.2 x108
-4
t
5.0 10 s
s
b. If the mirrors were 15 000 meters apart, how many rotations per second would the
revolving mirror have to make to obtain a clear image?
1
t   5  104 s   2.5  104 s
2
1
1
rps 

 400rps
number of faces  t 10   2.5 104 s 
therefore c 