Preparation problems

13-1. A car is headed north at 70 km/h. Give the magnitude and direction of the angular velocity of its 62-cm diameter wheels.

Solution:

We know that the velocity of a point r from the center of a body rotating with angular velocity w is given by:

vt = wr
with the direction determined by the right-hand rule. Since the tire is traveling north, the angular velocity should point in the west direction (it should always point 90° to the left of the car). We can then find the magnitude by:
w
=
vt
r
=
70  km/h · 1000  m/km
3600  s/h
0.31  m
w
=
62.8  rad/s west
13-2. If the car of Problem 13-1 makes a 90° left turn lasting 25 s, determine the angular acceleration of the wheels.

Solution:

We know that the angular velocity of the wheels is 62.8 rad/s, and that the direction should always be to the left of the car's direction of travel. Let's define a coordinate system with north as j and east as i. Then the angular velocity vector starts out as wi = -62.8 rad/s i, and finishes as wi = -62.8 rad/s j.

Using our definition of angular acceleration, we should be able to solve the problem. We recall

®
a
 
 =
d ®
w
 
dt
Here we are in a situation where we want the average values, so we can replace the derivative with the changes in value ( d ®D). This gives us
®
a
 
=
D ®
w
 
Dt
=
®
wf
 
 - ®
wi
 
Dt
=
(-62.8  rad/s   ^
i
 
 ) - (-62.8  rad/s   ^
j
 
 )
25  s
®
a
 
=
-2.50   ^
i
 
 -2.50   ^
j
 
  rad/s2

13-6. A rod is free to pivot about one end, and a force F is applied at the other end, as shown in Fig. 13-26 (p. 316). What are the magnitude and direction of the torque about the pivot point?

Solution:

We can use our simple definition of torque to find its magnitude, and the right hand rule should give us the direction. First lets find the direction. Placing the fingers of your right hand along the radius vector from the pivot point to the point at which the force is applied, wrap your fingers into the direction of the force. Our thumb should point in the direction of the torque, which in this case is into the page.

Now lets find the magnitude. This is given by

ê
ê 		®
t
 
 ê
ê 		= ê
ê 		®
r
 
 × ®
F
 
 ê
ê 		= rFsinq
Plugging in our values, we see
t
=
(0.8  m)(5  N) sin(20°)
t
=
1.36  N ·m  into  the  page



Follow-up problems

12-23. A 1.5-m diameter wheel is mounted on an axle ghrough its center. (a) Find the net torque about the axle due to the forces shown in Fig. 12-38 (on p. 296). (b) Are there any forces that don't contribute to the torque?

Solution:

(a) We will use our simple definition of torque

t = rF sinq
where r is the distance between the point at which the force is applied and the axis of rotation, F is the magnitude of the force applied, and theta is the angle between the force and the radius vector to the point of application. This will allow us to calculate the magnitude of each force (remembering to use the right hand rule to determine the sign of each torque). Adding all of the torques will give us the net torque, meaning it is given by
tnet =
å
 i 
 ri Fi sinqi
Plugging in our numbers, we see
tnet
=
r1 F1 sinq1 + r2 F2 sinq2 + r3 F3 sinq3 + r4 F4 sinq4 + r5 F5 sinq5
=
(0.75  m)(1.5  N) sin(0°) + (0.25  m)(3.0  N) sin(45°) + (0.5  m)(2.0  N) sin(270°)
+ (0.75  m)(1.8  N) sin(60°) + (0  m)(2.5  N) sin(90°)
tnet
=
0.699 N ·m

(b) It is obvious when looking at the equations above that F1 and F5 do not contribute to the net torque.

12-24. Four equal masses m are located at the corners of a square of side l, connected by essentially massless rods. Find the rotational inertia of this system about an axis (a) that coincides with one side and (b) that bisects two opposite sides

Solution:



(a) The rotational inertia of a collection of points is given by
I =
å
 i 
 mi ri2
where mi is the mass of each object and ri is the distance between that object and the axis of rotation. Looking at the picture above and numbering the masses clockwise from the top left corner, we get
I
=
m ·(0)2 + m ·l2 + m ·l2 + m ·(0)2
I
=
2ml2

(b) Again, the rotational inertia of a collection of points is given by

I =
å
 i 
 mi ri2
Remembering that ri is the distance between that object and the axis of rotation (labeled (b) in the picture), and numbering the masses clockwise from the top left corner, we see
I
=
m ·(l/2)2 + m ·(l/2)2 + m ·(l/2)2 + m ·(l/2)2
I
=
ml2

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Solutions translated from TEX by TTH, version 1.57.