This document discusses uniform circular motion and related concepts. It begins by defining uniform circular motion as motion at constant speed in a circular path. It then derives the formula for centripetal acceleration and explains that a centripetal force is needed to provide the acceleration toward the center required for circular motion. Examples are provided to illustrate calculating centripetal force for different objects in circular motion, including effects of speed and radius. The document also discusses banked curves and satellites in circular orbits, providing the relevant equations and example calculations.

1. Chapter 5
Dynamics of Uniform
Circular Motion

2. Chapter 5
Dynamics of Uniform
Circular Motion
Tutorial Problems Chapter 5
1, 4, 15, 18, 21, 23, 24, 25, 27, 28, 30

3. Topic for Discussion: Speed
Limit around a curve
1. What is the meaning of the sign indicated?
2. How did they come to say the speed indicated should be 100 km/h?.
3. Do you think the small car and the big truck or bus should obey the
same speed limit when passing through the indicated sign of 100 km/h
4. What do you think can happen if one travels more or less than the
indicated speed limit?

4. A ball is thrown vertically upwards from the surface of the
earth. Consider the following quantities based on the motion
of the ball.
1: Speed; 2: velocity; 3: acceleration
Speed, velocity and acceleration activities 5 02 2015
1.1 Which of these is (are) zero when the ball has reached the
maximum height at position B? Select the correct one
A: 1 and 2 only, B: 1 and 3 only, C: 1 only, D: 2 only, E: 1, 2 and 3
1.2 What do you think will be the magnitude and direction of acceleration
of an object at the following points? Give reasons for your answer.
1.2.1 Point A moving up magnitude_____direction_______(UP or DOWN)
1.2.2 Point B at maximum height: magnitude___direction__(UP or DOWN)
1.2.3 Point C moving down: magnitude___direction_____ (UP or DOWN)

5. A ball is thrown vertically upwards from the surface of the
earth. Consider the following quantities based on the motion
of the ball.
1: Speed; 2: velocity; 3: acceleration
Speed, velocity and acceleration activities 9 02 2015
1.3 What do you think will happen to the velocities of an object at the
following points? (Only write, increase, decrease, zero or remains the
same and give reason(s))
1.3.1 At position A : increase, decrease, zero or remains the same
1.3.2 At Position B: increase, decrease, zero or remains the same
1.3.3 At position C: increase, decrease, zero or remains the same
What about the accelerations at the points above?
1.2.1 At position A : increase, decrease, zero or remains the same
1.2.2 At Position B: increase, decrease, zero or remains the same
1.3.3 At position C: increase, decrease, zero or remains the same

6. Newton’s Laws of Motion
State and apply Newton’s Laws of Motion
Conditions for equilibrium: How do we know if objects are in equilibrium?
• Sum of the forces is equal to zero (along the x and y independently)
• Objects move at the same velocity
• The acceleration is zero
What is ACCELERATION?
Do you think an object moving at constant speed around a circular
track is accelerating?
Claim Evidence and reasoning format of an answer.

7. 5.1 Uniform Circular Motion
DEFINITION OF UNIFORM CIRCULAR MOTION
Uniform circular motion is the motion of an object
traveling at a constant speed on a circular path.

8. 5.1 Uniform Circular Motion
Let T be the time it takes for the object to
travel once around the circle.
v
r
T

2r
What is the formula to calculate the circumference of the circle?
Then the speed v around the circular path will be given by
Do example 1 as a homework,
𝐶 = 2𝜋𝑟
R the radius of the circle and
T the period in seconds

9. 5.2 Centripetal Acceleration
In uniform circular motion, the speed is constant, but the
direction of the velocity vector is not constant.

90 

90
 

10. Similarities of Triangles (back 2000)
Two triangles are similar if:
(a) The corresponding angles are equal
(b) The corresponding sides are in proportion
Note:
-Corresponding sides are sides opposite equal angles
-Corresponding angles are angles opposite sides in proportion

11. Your Turn
Consider ∆𝐶𝐵𝐻 𝑎𝑛𝑑∆𝐴𝐵𝐶 Prove that the two triangles are similar
What are equal angles in the two triangles?
What are the sides in proportion between the two triangles?
Lastly we derive the formula for centripetal acceleration using mathematical
geometry

12. 5.2 Centripetal Acceleration
r
tv
v
v 


r
v
t
v 2



r
v
ac
2

Diagram a and b (sides in proportion are equal)
Centripetal acceleration

13. 5.2 Centripetal Acceleration
The direction of the centripetal acceleration is toward the
center of the circle; in the same direction as the change in
velocity.
r
v
ac
2

Note that the velocity is constant and only time is changing

14. Your Turn
Which Way Will the Object Go?
An object is in uniform circular motion. At point O it
is released from its circular path. Does the object
move along the straight path between O and A or
along the circular arc between points O and P ?
Claim:
Evidence:
Reasoning:

15. The Impact of Radius on Centripetal Acceleration
The bobsled track contains turns with radii of 33 m and
24 m. Without doing any calculation which of the curve
will have greater acceleration?
Claim:
Evidence:
Reasoning:
Justifications using calculations
  gac 6.3sm35
m33
sm34 2
2

  gac 9.4sm48
m24
sm34 2
2


16. 5.3 Centripetal Force
  aF

m
m

F
a


Recall Newton’s Second Law
When a net external force acts on an object
of mass m, the acceleration that results is
directly proportional to the net force
and has a magnitude that is inversely
proportional to the mass.
The direction of the acceleration is
the same as the direction of the net force.

17. 5.3 Centripetal Force
Thus, in uniform circular motion there must be a net
force to produce the centripetal acceleration.
The centripetal force is the name given to the net force
required to keep an object moving on a circular path.
The direction of the centripetal force always points toward
the center of the circle and continually changes direction
as the object moves.
r
v
mmaF cc
2


18. 5.3 Centripetal Force
Example 5: The Effect of Speed on Centripetal Force
The model airplane has a mass of 0.90 kg and moves at
constant speed on a circle that is parallel to the ground.
The path of the airplane and the guideline lie in the same
horizontal plane because the weight of the plane is balanced
by the lift generated by its wings. Find the tension in the 17 m
guideline for a speed of 19 m/s.
r
v
mTFc
2

   N19
m17
sm19
kg90.0
2
T

19. 5.3 Centripetal Force
Check Conceptual Example 6: A Trapeze Act
In a circus, a man hangs upside down from a trapeze, legs
bent over and arms downward, holding his partner. Is it
harder for the man to hold his partner when the partner
hangs straight down and is stationary of when the partner
is swinging through the straight-down position?

20. Personal Activity: Due: 23 Feb
2015
To negotiate an unbanked curve at faster
speed, a driver puts a couple of sand bags
in his van to increase the force of friction
between the tires and the road. Will the
sand bags help?
Claim: (1)
Scientific Evidence: (2)
Reasoning: (2)

21. 5.4 Banked Curves
Note:
On an unbanked curve, the static frictional force
provides the centripetal force. Centripetal force and
Safe Driving Check example page 141

22. Highway Curves: Banked & Unbanked
Case 1 - Unbanked Curve:
When a car rounds a curve, there MUST be a net force toward the circle
center (a Centripetal Force) of which the curve is an arc.
If there weren’t such a force, the car couldn’t follow the curve, but would (by
Newton’s 1st Law) go in a straight line.
On a flat road, this Centripetal Force is the static friction force.
No static friction?
 No Centripetal Force
 The Car goes straight!
=

23. Note:
If the friction force isn’t sufficient, the car will tend to move
more nearly in a straight line (Newton’s 1st Law) as the skid
marks show.
As long as the tires don’t slip, the
friction is static. If the tires start
to slip, the friction is kinetic,
which is bad
1. The kinetic friction force is
smaller than the static friction
force.
2. The static friction force can point
toward the circle center, but the
kinetic friction force opposes the
direction of motion, making it
difficult to regain control of the car
& continue around the curve.

24. 5.4 Banked Curves
On a frictionless banked curve,
• the centripetal force is the horizontal component of
the normal force.
• The vertical component of the normal force balances the
car’s weight.

25. 5.4 Banked Curves
r
v
mFF Nc
2
sin   )1(cos mgFN 
)2(sin
2
r
v
mFN 
:)2()1( bydivide
rg
v2
tan 
Conclusion:
The speed around the curve
Is independent of the mass of
the car or bus

26. 5.4 Banked Curves
r
v
mFN
2
sin 
mgFN cos
rg
v2
tan 

27. 5.4 Banked Curves
Example 8: The Daytona 500
The turns at the Daytona International Speedway have a
maximum radius of 316 m and are steely banked at 31
degrees. Suppose these turns were frictionless. At what
speed would the cars have to travel around them?
rg
v2
tan  tanrgv 
    mph96sm4331tansm8.9m316 2
 
v

28. Satellites

29. 5.5 Satellites in Circular Orbits
There is only one speed that a satellite can have if the
satellite is to remain in an orbit with a fixed radius.
What are the names and formula of the forces acting ?

30. 5.5 Satellites in Circular Orbits
r
v
m
r
mM
GF E
c
2
2

r
GM
v E

Hence the speed that a satellite is independent of
its mass but its radius (r) from the Earth.

31. 5.5 Satellites in Circular Orbits
Example 9: Orbital Speed of the Hubble Space Telescope
Determine the speed of the Hubble Space Telescope orbiting
at a height of 598 km above the earth’s surface.
  
 hmi16900sm1056.7
m10598m1038.6
kg1098.5kgmN1067.6
3
36
242211





v

32. 5.5 Satellites in Circular Orbits
T
r
r
GM
v E 2

EGM
r
T
23
2


33. 5.5 Satellites in Circular Orbits
Global Positioning System
hours24T
EGM
r
T
23
2


34. 5.5 Satellites in Circular Orbits

35. Section 5.2 no 1 p 149
1. Two cars are traveling at the same constant speed v. Car A is
moving along a straight section of the road, while B is rounding a
circular turn. Which statement is true about the acceleration of the
cars?
(a) The acceleration of both cars is zero, since they
are traveling at a constant speed.
(b) Car A is accelerating, but car B is not accelerating.
(c) Car A is not accelerating, but car B is accelerating.
(d) Both cars are accelerating.
Explanation :The velocity of car A has a constant
magnitude (speed) and direction. Since its velocity is constant, car A
does not have an acceleration. The velocity of car B is continually
changing direction during the turn. Therefore, even though car B has a
constant speed, it has an acceleration (known as a centripetal
acceleration).

36. Section 5.4: Banked Curves
10. Two identical cars, one on the moon and one on the earth, have the same
speed and are rounding banked turns that have the same radius r. There are
two forces acting on each car, its weight mg and the normal force exerted by
the road. Recall that the weight of an object on the moon is about one-sixth of
its weight on earth. How does the centripetal force on the moon compare with
that on the earth?
(a) The centripetal forces are the same.
(b) The centripetal force on the moon is less than that on the earth.
(c) The centripetal force on the moon is greater than that on the earth.
Explanation
The magnitude of the centripetal force is given by Fc = mv2
/r. The two cars have the
same speed v and the radius r of the turn is the same. The cars also have the same
mass m, even though they have different weights due to the different
accelerations due to gravity. Therefore, the centripetal accelerations are the same.

37. Section 5.5: Satellites in circular orbirts
11. Two identical satellites are in orbit about the earth. One orbit has a
radius r and the other 2r. The centripetal force on the satellite in the
larger orbit is ________ as that on the satellite in the smaller orbit.
(a) the same
(b) twice as great
(c) four times as great
(d) half as great
(e) one- fourth as great
Explanation:
(e) The centripetal force acting on a satellite is provided by the gravitational force.
The magnitude of the gravitational force is inversely proportional to the radius
squared (1/r
2
), so if the radius is doubled, the gravitational force is one fourth as great;
1/2
2
= 1/4.

38. Question: Skidding on a curve
A car, mass m = 1,000 kg car rounds a curve on a flat road of radius r = 50 m at a constant
speed v = 14 m/s (50 km/h). Will the car follow the curve, or will it skid? Assume
a. Dry pavement with the coefficient of static friction μs = 0.6.
b. Icy pavement with μs = 0.25.
Free Body
Diagram
Approach
Step 2:
Draw the FBD and identify all forces acting on the car
 The force of gravity 𝐹𝐺 = 𝑚𝑔 = 1000 9.8 = 9800 𝑁 downward
 The normal force 𝐹 𝑁 = 𝑚𝑔
 The horizontal frictional force 𝐹𝑓 = 𝜇 𝑠 𝐹 𝑁 = 𝜇 𝑠 𝑚𝑔
Step 1: Criteria to follow the curve or skidding.
Reasoning: The car will follow the curve if the maximum static frictional force is
greater than the centripetal force.
That is 𝐹𝑓 𝑚𝑎𝑥
≥ 𝐹𝐶 where 𝐹𝐶= 𝑚
𝑣2
𝑟
= 1000
(14)2
50
= 3900 𝑁
Noting that once the car skid, the static frictional force becomes kinetic
frictional force which is less than static frictional force
Answer: Compare the static frictional force with the centripetal force
If 𝑭 𝒇 𝒎𝒂𝒙
> 𝑭 𝑪 hence the car will follow the curve
If 𝐹𝑓 𝑚𝑎𝑥
< 𝐹𝐶 hence the car will skid
a. Dry pavement with the coefficient of static friction μs = 0.6
𝐹𝑓 = 𝐹𝑓 = 𝜇 𝑠 𝐹 𝑁 = (0.6)(9800) = 5900 𝑁 > 𝐹𝐶 𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑐𝑎𝑟 𝑤𝑖𝑙𝑙 𝑓𝑜𝑙𝑙𝑜𝑤 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒
b. Icy pavement with μs = 0.25
𝐹𝑓 = 𝐹𝑓 = 𝜇 𝑠 𝐹 𝑁 = (0.25)(9800) = 2500 𝑁 < 𝐹𝐶 𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑐𝑎𝑟 𝑤𝑖𝑙𝑙 𝑠𝑘𝑖𝑑

39. Vertical circular motion
Question: Is vertical circular motion uniform? Explain
• When the speed of travel on a circular path changes from moment to moment, the motion is said
to be non-uniform.
• But, we can use the concepts that apply to uniform circular motion to gain considerable insight into
the motion that occurs on a vertical circle.
Considering the second diagram
There are four points on a vertical circle where the centripetal force can be identified easily, denoted by points 1 to 4
• Keep in mind that the centripetal force is not a new and separate force of nature. Instead, at each point the centripetal force is
the net sum of all the force components oriented along the radial direction, and it points toward the centre of the circle.
• The drawing shows only the weight of the cycle plus rider (magnitude = mg) and the normal force pushing on
the cycle (magnitude = FN).
• The magnitude of the centripetal force at each of the four points is given as follows in terms of mg and FN:
• As the cycle goes around, the magnitude of the normal force changes. It changes because the speed changes
and because the weight does not have the same effect at every point
Case 1:
At the bottom, the normal force and the weight opposes one another, giving a resulting centripetal force of magnitude FN1 - mg and the
second law becomes: Resultant F: 𝐹 𝑁1 − 𝑚𝑔 = 𝑚
𝑣1
2
𝑟
Case 3:
On top, the normal force and the weight are facing the same direction, giving a resulting centripetal force of magnitude
FN1 +mg and hence the second law becomes: 𝐹 𝑁3 + 𝑚𝑔 = 𝑚
𝑣3
2
𝑟
Cases 2 and 4:
In both cases, the normal force and weight are perpendicular to each other,, and weight not in the same direction as the centripetal
force hence the resulting centripetal force will be caused by the normal force:
second law becomes: 𝐹 𝑁2 = 𝑚
𝑣2
2
𝑟
and 𝐹 𝑁4 = 𝑚
𝑣4
2
𝑟

40. The case of the minimum force to keep an object at circular on top of vertical track
• Riders who perform the loop-the-loop trick must have at least a minimum speed at the top of the circle to remain on
the track. This speed can be determined by considering the centripetal force at point 3.
• The speed 𝑣3 in the equation 𝐹 𝑁3 + 𝑚𝑔 = 𝑚
𝑣3
2
𝑟
is a minimum when 𝑭 𝑵𝟑 is zero. Hence it will be given by
• Then, the speed is given by 𝒗 𝟑 = 𝒓𝒈 .
• At this speed, the track does not exert a normal force to keep the cycle on the circle, because the weight mg provides
all the centripetal force.
mg
FN
Free-body diagram of
the motorcycle
Do no problem no 46 on your own

41. 47. REASONING Because the crest of the hill is a circular arc, the motorcycle’s
speed v is related to the centripetal force Fc acting on the motorcycle: 2
cF mv r
where m is the mass of the motorcycle and r is the radius of the circular
crest. Solving Equation 5.3 for the speed, we obtain 2
cv F r m or cv F r m .
The free-body diagram shows that two vertical forces act on the motorcycle. One is
the weight mg of the motorcycle, which points downward. The other is the normal
force FN exerted by the road. The normal force points directly opposite the
motorcycle’s weight.
mg
FN
Free-body diagram of
the motorcycle
Note that the motorcycle’s weight must be greater than the normal force. The reason
for this is that the centripetal force is the net force produced by mg and FN and
must point toward the center of the circle, which lies below the motorcycle.
Only if the magnitude mg of the weight exceeds the magnitude FN of the normal
force will the centripetal force point downward.
Therefore, we can express the magnitude of the centripetal force as Fc = mg − FN.
With this identity, the relation cv F r m becomes
When the motorcycle rides over the crest sufficiently fast, it loses contact with the
road. At that point, the normal force FN is zero. In that case, Equation (1) yields
the motorcycle’s maximum speed:

42. REASONING
The drawing at the right shows the two forces that act
on a piece of clothing just before it loses contact with
the wall of the cylinder.
At that instant the centripetal force is provided by
the normal force 𝐹 𝑁 and the radial component of the
weight.
From the drawing, the radial component of the weight
is given by
mg mg mgcos = cos (90 – ) = sin  
Therefore, with inward taken as the positive direction, Equation 5.3 ( Fc  mv2
/r )
gives

43. Therefore, with inward taken as the positive direction, Equation 5.3 ( Fc  mv2
/r )
gives
F mg
mv
r
2
N
sin = 
At the instant that a piece of clothing loses contact with the surface of the drum,
FN
 0N, and the above expression becomes
mg
mv
r
2
sin =
Substituting the equation of speed, v  2r /T, we get
g
r T
r
r
T
2
sin =
 ( / )2 4 2
2

Solving for T and substituting unknowns we get
T
r
g
r
g
  


4
2 1 17
2





sin sin
2
0.32 m
9.80 m / s sin 70.0
s2
c h .
Therefore, the number of revolutions per
second that the cylinder should make is 1 1
1 17T
 
. s
0.85 rev / s

44. Extra Problems
REASONING
The speed v of a satellite in circular orbit about the earth is given by 𝑣 =
𝐺𝑀 𝐸
𝑟
, where G is the universal gravitational
constant, and 𝑀 𝐸 is the mass of the earth, and r is the radius of the orbit.
The radius is measured from the center of the earth, not the surface of the earth, to the satellite.
Therefore, the radius is found by adding the height of the satellite above the surface of the earth to the radius of the
earth (6.38 × 106
𝑚).
Satellite A
Satellite B
𝑉 = 7690 𝑚/𝑠 𝑉 = 7500 𝑚/𝑠

45. Vertical Circular Motion
For vertical circular motion, the motion is not uniform, as the object increases
speed on the downward swing and decreases on the way up. When analysing
vertical circular motion, one finds that there are two forces acting on the object.
These two forces are T, the tension in the
string, and FW, the weight of the object, as
shown below.
The weight, FW, can be resolved into a tangential component
expressed as FWsinθ and a radial component expressed as FWcosθ
𝑎 𝑇 =
𝐹 𝑇
𝑚
=
𝐹 𝑊 𝑠𝑖𝑛𝜃
𝑚
=
𝑚𝑔𝑠𝑖𝑛𝜃
𝑚
= 𝑔𝑠𝑖𝑛𝜃
𝑎 𝑅 =
𝐹 𝑅
𝑚
=
𝑇 − 𝐹 𝑊 𝑐𝑜𝑠𝜃
𝑚
=
𝑇 − 𝑚𝑔𝑐𝑜𝑠𝜃
𝑚
=
𝑇 − 𝑚𝑔𝑐𝑜𝑠𝜃
𝑚
Applying Newton's second law to express the tangential
acceleration gives:
Similarly, the radial acceleration can be expressed as:
Substituting
𝑣2
𝑟
for 𝑎 𝑅 and solving for T yields: 𝑇 = 𝑚(
𝑣2
𝑟
+ 𝑔𝑐𝑜𝑠𝜃)
Two interesting
points to
consider are the
top and the
bottom of the
circle.
At the lowest point (Bottom) of the circular
path, θ = 0° and cos 0° = 1. Substituting into
the equation for T yields:
𝑇 = 𝑚(
𝑣2
𝑟
+ 𝑔)
At the highest point of the circular path,
θ = 180° and cos 180° = -1. Substituting
into the equation for T yields:
T = m(
𝑣2
𝑟
- g)

46. At Gold Reef City, a child of mass m rides on a Ferris wheel as
shown in the diagram above. The child moves in a vertical
circle of radius 10.0 m at a constant speed of 3.00 m/s.
(a) Calculate the force exerted by the seat on the child at the
bottom of the ride. Express your answer in terms of the
weight of the child, mg.
(b) Calculate the force exerted by the seat on the child at the
top of the ride.
Test Your Understanding

47. (a) At the bottom, Consider the forces acting at the bottom: Force of gravity
𝐹𝑔 = 𝑚𝑔, downwards and the normal force 𝐹 𝑁 upwards. Hence
𝑁𝑒𝑤𝑡𝑜𝑛′
𝑠 𝑠𝑒𝑐𝑜𝑛𝑑 𝑙𝑎𝑤:
𝐹 = 𝐹 𝑁 − 𝑚𝑔 = 𝑚
𝑣2
𝑟
𝐹 𝑁 = 𝑚𝑔 + 𝑚
𝑣2
𝑟
= 𝑚(𝑔 +
𝑣2
𝑟
)
𝐹 𝑁 = 𝑚𝑔 1 +
𝑣2
𝑔𝑟
= 1.09 𝑚𝑔
(b) At the bottom, Consider the forces acting at the bottom: Force of gravity
𝐹𝑔 = 𝑚𝑔, downwards and the normal force 𝐹 𝑁 is always upwards.
Because the child is always in upright position.
𝐹 = 𝑚𝑔 − 𝐹 𝑁 = 𝑚
𝑣2
𝑟
𝐹 𝑁 = 𝑚𝑔 − 𝑚
𝑣2
𝑟
= 𝑚(𝑔 −
𝑣2
𝑟
)
𝐹 𝑁 = 𝑚𝑔 1 −
𝑣2
𝑔𝑟
= 0.908 𝑚𝑔

48. Test your understanding:
An object with a mass of 8.0 kg is swung in a vertical circle of radius
2.4 m with a speed of 6.0 m/s.
(a) Determine the maximum and minimum tension in the string.
(b) The string breaks when the tension exceeds 340 N.
Determine the maximum speed of the object and where the
mass will be when the string breaks. Justify your answer as to
where the mass will be.
For solution check: http://mmsphyschem.com/verCMAns.htm

49. 1. SOLUTION Since 2
c
/a v r and  2 /v r T , the magnitude of the car’s
centripetal acceleration is
 
 
2
2 32 2
2
2c 2
2
4 10 m4
0.86 m/s
400 s
r
v rT
a
r r T


 
       
4 SOLUTION Using Equation 5.2 for the centripetal acceleration of each boat,
we have
2 2
A B
cA cB
A B
and
v v
a a
r r
 
Setting the two centripetal accelerations equal gives
2 2
A B
A B
v v
r r

Solving for the ratio of the speeds gives A A
B B
80 m
0.58
240 m
v r
v r
  
15. REASONING At the maximum speed, the maximum centripetal force acts on the
tires, and static friction supplies it. The magnitude of the maximum force of static
friction is specified by Equation 4.7 as MAX
s s Nf F , where s is the coefficient
of static friction and FN is the magnitude of the normal force. Our strategy, then,
is to find the normal force, substitute it into the expression for the maximum
frictional force, and then equate the result to the centripetal force, which is
2
c /F mv r , according to Equation 5.3. This will lead us to an expression for the
maximum speed that we can apply to each car.
SOLUTION Since neither car accelerates in the vertical direction, we can
conclude that the car’s weight mg is balanced by the normal force, so FN = mg.
From Equations 4.7 and 5.3 it follows that
2
MAX
s s N s c
mv
f F mg F
r
    
Thus, we find that
2
s sor
mv
mg v gr
r
  
Applying this result to car A and car B gives
A s, A B s, Bandv gr v gr  
Some of the Tutorial Solutions

50. In these two equations, the radius r does not have a subscript, since the radius is the
same for either car. Dividing the two equations and noting that the terms g and r
are eliminated algebraically, we see that
 s, B s, B s, BB
B A
A s, As, A s, A
0.85
or 25 m/s 22 m/s
1.1
grv
v v
v gr
  
 
    
18. REASONING The centripetal force Fc that keeps the car (mass = m, speed = v) on
the curve (radius = r) is 2
c /F mv r (Equation 5.3). The maximum force of static
friction MAX
sF provides this centripetal force. Thus, we know that MAX 2
s /F mv r
, which can be solved for the speed to show that MAX
s /v rF m . We can apply
this result to both the dry road and the wet road and, in so doing, obtain the desired
wet-road speed.
Applying the expression MAX
s /v rF m to both road conditions gives
MAX MAX
s, dry s, wet
dry wetand
rF rF
v v
m m
 
We divide the two equations in order to eliminate the unknown mass m and unknown
radius r algebraically, and we remember that
MAX MAX1
s, wet s, dry3
F F :
MAX
MAX
MAXMAX
s, wet
s, wetwet
dry s, drys, dry
1
3
rF
Fv m
v FrF
m
  
Solving for vwet, we obtain
dry
wet
25 m / s
14 m / s
3 3
v
v   

51. What you should do
Tutorial Problems Chapter 5
1, 4, 15, 18, 21, 23, 24, 25, 27, 28, 30, 35, 38, 55