Chapter 5
Uniform Circular Motion
First, radian and angular speed will be discussed.
Radian, the SI unit of Angle:
One radian of angle is the central angle in any circle which opposite arc is equal to the radius of that circle. If a piece of string is cut equal to the radius R of a circle and then placed on the edge of that circle, as shown, the central angle corresponding (or opposite) to that arc is called one "radian."
If a string, cut to length R,
is laid down on the edge 

Example 1: Naming 3.14rd as " π ", calculate angles 360^{o}, 180^{o}, 90^{o}, 60^{o}, 45^{o}, and 30^{o} in terms of π.
Solution:
Fraction:  1 full circle  1/2 circle  1/4 circle  1/6 circle  1/8 circle  1/12 circle 
Degrees:  360^{o}  180^{o}  90^{o}  60^{o}  45^{o}  30^{o} 
Radians:  2π = 6.28rd  π  π/2  π/3  π/4  π/6 
The Arc LengthCentral Angle Formula:
There is an easy formula that relates any central angle θ to its opposite arc s and the radius R of a circle. This formula is valid only if the central angle is measured or expressed in radians.
Example 2: Referring to the above figure, suppose angle θ is 148^{o} and R = 1.25 in. Calculate the length of arc AB.
Solution: Using S = R θ, and converting degrees to radians, yields: S = (1.25 in.)(148^{o})( 3.14rd / 180^{o} ) = 3.23 in.
The Angular Speed (ω):
The symbol "ω" is the lower case of the Greek letter Ω pronounced "omega." Angular speed ω is defined as the change in angle per unit of time. Mathematically, it may be written as
ω = Δθ/Δt. The preferred unit for angular speed is rd/s. The industrial unit for angular speed is rpm or revolutions per minute.
Note that each revolution or turn is 6.28 radians.
Example 3: A car tire spins at 240rpm. Calculate (a) its angular speed in rd /s, (b) the angle that any of its radii sweeps in 44 seconds, and (c) the arclength that any point on its outer edge travels during this time knowing that R = 14 inches. Make sure that you completely solve this problem on paper using horizontal fraction bars everywhere.
Solution: 240rpm is the angular speed, ω. All we need to do in part (a) is to convert it from rpm to rd/s.
(a) ω = 240 (rev/min) = 240[6.28 rd/60s] = 25 rd/s. Note: 1rev = 6.28rd and 1min = 60s.
(b) ω = Δθ/Δt ; Δθ = ωΔt ; Δθ = [25 rd/s] (44s) ; Δθ = 1100 rd.
(c) s = Rθ ; s = 14 in.(1100 rd) = 15400 in. = 1283 ft = 0.24 miles.
The Linear Speed  Angular Speed Formula:
The same way that S and θ are related, we can develop a formula that relates v to ω. The formula is: v = Rω.
Writing these two similar formulas together helps their recall: s = Rθ ; v = Rω.
The derivation is easy. All you need to do is to first write s = Rθ as Δs = RΔθ and then divide both sides of it by Δt.
We may reason that if s = Rθ, then Δs = RΔθ. Dividing through by Δt, results in:
[Δs /Δt] = R [Δθ/Δt] (*)
As we see Δs/Δt is v or the linear speed. Δθ/Δt is ω or the angular speed.
V = Rω. 
Note: If you think Equation (*) is wrong because of division through by Δθ, refer to the end of the chapter for details.
Example 4: The radius of a car tire is 14 inches. Calculate (a) the linear speed v of any point on its outer edge if the tire spins at a constant angular speed of 25rd/s. (b) Find the linear distance or the arc length that any such point travels in 44s.
Solution: (a) Using v = Rω yields: V = (14 in.)(25 rd/sec) = 350 in/s.
(b) s = vt ; s = (350 in./s)(44s) = 15400in. Use horizontal fraction bars when you solve. Note: In Part (b), the equation x = (1/2)at^{2} + v_{i} t is used in which a = 0 due to constant linear speed and x is replaced by s. We think of the arc length as a long string that is initially wrapped around the tire and as the tire rolls on the road, the string unwraps leaving the string on the road like a straight line (the distance traveled) for which the equation s = Vt or x = Vt is valid.
Chapter 5 Test Yourself 1:
1) Radian is a unit of (a) length (b) angle (c) area. click here.
2) A central angle is an angle that has its vertex at the center of a (a) triangle (b) square (c) circle.
3) In any circle, the size of any central angle is equal to (a) the arc opposite to it when expressed in radians (b) half of the arc opposite to it (c) the radius of that circle.
4) Draw a circle and select two central angles in it, one equal to 90^{o} and one equal to 45^{o}. Since 90^{o} is 1/4 of 360^{o}, verify that the arclength opposite to the 90^{o}angle you chose is also 1/4 of the whole circle. Also, verify that the 45^{o}angle you chose is opposite to an arc that is exactly 1/8 of the whole circle. What conclusion do you draw? Write it down. Is your conclusion in line with the correct answer to Question 3? click here.
5) 1rd is the central angle which opposite arclength equals (a) the radius of the circle (b) the diameter of the circle (c) the perimeter of the circle.
6) The central angle which opposite arc equals the perimeter of the circle is (a) 360^{o} (b) 2π radians (c) both a and b.
7) The central angle which opposite arc equals half of the circle is (a) 120^{o} (b) 180^{o} (c) 3.14radians (d) both b and c.
8) 2.00 radians of angle is equivalent to (a) 114.6^{o} (b) 120^{o} (c) 170^{o}. click here.
9) S = Rθ calculates the arc length that is opposite to central angle θ only if θ is expressed in (a) degrees (b) radians (c) grads.
10) The arc length opposite to a 2.00rd central angle in a circle of radius R = 6.00" is (a) 12.0" (b) 8.00" (c) 18.0".
11) The arc length opposite to a 114.6^{o} central angle in a circle of radius R = 6.00" is (a) 18.0" (b) 28.0" (c) 12.0".
12) Angular speed is defined as (a) the arc length traveled per unit of time (b) the angle traveled per unit of time (c) the angle swept by by any one radius per unit of time (d) both b and c. click here.
13) If angular change is Δθ and time change is Δt, then angular speed, ω, is (a) Δθ /Δt (b) Δt /Δθ (c) ΔθΔt.
14) The angular speed formula ω = Δθ /Δt is the counterpart of the linear speed formula v = Δx /Δt. (True) or (False)?
15) A spoke on a bicycle wheel travels or sweeps 150rd of angle every minute. Its angular speed is (a) 150rd/s (b) 150rd/min (c) 2.50rd/s (d) both b and c. click here.
16) RPM is (a) a unit of angle (b) a unit of angular acceleration (c) an industrial unit for angular speed.
17) rd/s is (a) an industrial unit for angle (b) the SI unit for angular speed (c) neither a nor b. click here.
18) 180 rpm is the same thing as (a) 3.0 rps (b) 3.0 revolutions per second (c) 3.0 turns per second (d) a, b, and c.
19) 240 rpm is the same thing as (a) 240 turns per minute (b) 4.0 turns per second (c) both a and b.
20) 1 rpm is the same thing as 6.28 rd/min. True or False? click here.
21) 3600 rpm is equivalent to (a) 60 rev/s (b) 60x6.28rd /s (c) 377 rd/s (d) a, b, and c.
22) A helicopter propeller rotates at 956 rpm. Its angular speed is (a) 100 rd/s (b) 200 rd/s (c) 300 rd/s. click here.
23) The formula that relates linear speed to angular speed is (a) s = Rθ (b) v = Rω (c) x = vt + 1/2 at^{2}.
24) A helicopter propeller rotates at 956 rpm. The linear speed of the tip of its propeller that is 5.00m long is (a) 500m/s (b) 750m/s (c) 956m/s. click here.
25) A helicopter propeller rotates at 956 rpm. The linear speed of the midpoint of its propeller that is 2.50m from its axis of rotation is (a) 500m/s (b) 750m/s (c) 250m/s.
26) A helicopter propeller rotates at 956 rpm. The angular speed of the midpoint of it is (a) the same as the angular speed of the tip of it (b) is 1/2 of the angular speed of the tip of it (c) neither a nor b. click here.
27) All points on a rotating solid wheel have the same (a) linear speed (b) acceleration (c) angular speed.
28) The linear speed of a point on a solid rotating disk (a) depends on R, its distance from the center of rotation (b) does not depend on R (c) depends on the angular speed of the disk. (d) both a and c.
29) All points on the outer edge of a rotating solid wheel have (a) the same angular speed (b) the same linear speed (c) the same linear velocity (d) a and b. click here.
30) A solid wheel of radius 0.400m spins at 478 rpm. Find its angular speed in rd/s as well as the linear speed of points on it that are at radii 0.100m, 0.200m, and 0.300m.
The Uniform Circular Motion
When a particle of mass M moves along a circular path at a constant rpm (revolutions per minute), its motion is called the "uniform circular motion."
Period of Rotation (T): The time it takes for the particle (mass M) to make one full turn is called the "period of rotation." T is the same thing as "seconds per turn."
Frequency (f): The number of turns per unit of time (usually per second) is called the frequency of rotation. f is the same thing as "turns per second."
From the definitions of T and f, it is easy to understand that the two are reciprocals.
Example 5: A wheel is turning at 2 turns per second. Calculate the time for each turn or simply its period.
Solution: It is clear that each turn is made in (1/2)s; In other words, T = (1/2)s. This is the same thing as using the formula T=1/f and replacing (f) by 2 per second. If the frequency (f) is 3 per second, the time of each turn or (T) is T = (1/3)s.
The Relation Between ω and f :
The following formula relates ω to f. Since f is the # of turns per second, ω the # of radians per second, and each turn = 2π radians, we get:
ω = 2πf or, 
# of rd/s = 2π [# of turns/s] 
Example 6: A car tire is spinning at a constant rate of 660rpm when the car's linear speed is 22.0m/s along the road. Determine (a) the frequency of rotation of the tire, (b) the period of rotation, and (c) the radius of the tire. Note: π = 3.14.
Solution:
(a) The angular speed is given: ω = 660rpm. Converting it to rd/s, we get:
ω = 660(rev/min) = 660 [2π rd/60s] = 69.1 rd/s.
From ω = 2πf ; f = ω/2π = [69.1 rd/s] / 2π = 11.0/s ; f = 11.0 Hz.
{Note: Of course, 660 turns per minute is like 11 turns per second.}
(b) Period: T = 1/f ; T = (1/11)s = 0.0909s.
(c) V = Rω ; R = V/ω ; R = [22.0m/s] / [69.1 rd/s] = 0.318m.
Example 7: As we know, the Earth goes around the Sun once per year. Calculate the linear speed of the Earth as it travels around the Sun. The average distance between the two is R = 150 million kilometers. Suppose the orbit is a perfect circle although it is not.
Solution: The angular speed of the Earth in its rotation around the Sun is:
ω = 1 rev/yr = 2π rd / [365 x 24 x 3600s] = 1.99x10^{7} rd/s.
V = Rω = 150x10^{6}x10^{3}m [1.99x10^{7} rd/s] = 30,000 m/s = 30 km/s = 19 mi/s.
Note that the linear speed is 19 miles per second, and NOT 19 miles per hour!
The Centripetal Force:
While in outer space (very far from planets and stars), if an astronaut throws a rock in any direction, the rock keeps moving in that direction along the initial velocity vector given to it. As long as there is no any other force to change the rock's direction, it will keep moving in that same direction. In general, an object tends to move along a straight line unless it is forced to do otherwise. If an object is moving along a circular path, it is because it is forced to do so. To force an object to take on a circular path, it must be constantly pushed or pulled toward the center of rotation. That is why such force is called the "centripetal force, F_{c}."
As an example, when you tie a rock to a string and spin it above your head in a horizontal plane, your hand is constantly pulling the rock toward your fingers that is the center of rotation. If you let go of the string, the centripetal force is eliminated and the rock will no longer follow that circular path; therefore, the cause of circular motion is centripetal force, a force that keeps pulling the object toward the center of rotation regardless of where the object is in its circular path. The magnitude of centripetal force is given by
where M is the mass of the circulating object, v its linear speed, and R its radius of rotation. As Particle M moves from Position 1 to Position 2, not only the direction of V changes from V_{1} to V_{2} , but also the direction of F_{c} changes from F_{c1} to F_{c2} . 

Example 8: A rock of mass 0.22kg is attached to a string of length 0.43m and is given a circular motion in a horizontal plane at a rate of 180rpm. Calculate the tension in the string. In other words, find the centripetal force that the string keeps exerting on the rock pulling it toward the center of rotation. Draw a circle and show M, V, R , and F_{c} on it as shown above.
Solution: ω = 180 rev/min = 180 (6.28 rd / 60s) = 18.84 rd/s. (Not rounded)
v = Rω ; v = (0.43m)(18.84 rd/s) = 8.1 m/s.
F_{c} = Mv^{2}/R ; F_{c} = 0.22kg (8.1m/s)^{2} / 0.43m = 34N.
Example 9: A 48.0cm string connected to a 56.0gram piece of lead is 24.0N strong. Calculate the maximum RPM at which the lead piece can be spun in a horizontal plane before the string breaks.
Solution: First we need to calculate the maximum linear speed v the lead piece can attain in its circular motion before the string breaks. Using F_{c} = MV^{2}/R and solving for V, we get:
V^{2} = RF_{c}/M = 0.480m(24.0N)/0.0560kg = 205.7 m^{2}/s^{2} ; V = 14.3m/s.
v = Rω ; ω = v/R ; ω = [14.3m/s] /0.480m = 29.9 rd/s.
Converting to RPM: ω = 29.9 rd/s = 29.9 [(1/6.28)rev.] / [(1/60) min.] = 285 RPM.
Chapter 5 Test Yourself 2:
1) In uniform circular motion, (a) the angular speed is constant (b) the linear speed is constant (c) the linear velocity is constant (d) the rpm is constant (e) a, b, and d. click here.
2) In uniform circular motion, the radius of rotation can be varying and the motion still be uniform. (a) True (b) False
3) In any curved motion, including uniform circular motion, the velocity vector is always tangent to the path of motion at any point. (a) True (b) False click here.
4) In uniform circular motion, the velocity vector at any point on the path is perpendicular to the radius of rotation at that point. (a) True (b) False
5) The period of rotation is the (a) time of 1rd of angle to be swept (b) time of completing 100 turns (c) time of completing one turn. click here.
6) Frequency, f, is the (a) # of turns per second (b) # of rotations per second (c) reciprocal of period, T (d) a, b, and c.
7) When f = 5s^{1}, it means that the spinning object makes 5 turns (a) per minute (b) per second (c) neither a, nor b.
8) When f = 5s^{1}, it is the same thing as writing (a) f = 5/s (b) f = 5 rev/s (c) f = 5 Hertz (d) f = 5Hz (e) all of a, b, c, and d.
9) f = 10.0 rev/s is the same thing as (a) f = 98.6 rd/s (b) 31.4 rd/s (c) f = 10.0s^{1} as well as ω = 62.8 rd/s. click here.
10) f is the rotation frequency, but ω is the angular frequency that means the # of radians per second. (a) True (b) False
11) The formula that relates frequency f (the # of rotations per second) to ω (the # of radians swept per second) is ω=2πf. (a) True (b) False click here.
12) A bicycle wheel makes 10.0 rotations per second.
We may write (a) f = 10.0s^{1} & ω = 62.8 /s (b) f = 10.0s^{1} & ω = 62.8 rd/s (c) f = 10.0s & ω = 62.8 rd/s.
13) A car tire runs at 480 rpm. We may write: click here.
(a) f =8.0s^{1} & ω=480/s (b )f =480s^{1} & ω=8.0 rd/s (c) f =8.0/s & ω=50.24 rd/s.
14) The angular speed, ω of a juicer is 314 rd/s. Its frequency, f is (a) 50.0/s (b) 50.0s^{1} (c) 50.0Hz (d) a, b, & c.
15) A culdesac has a radius of 60.0ft. Two straight lines from the center of the culdesac are drawn to its circular edge. The angle in between measures 36^{o}. The arc length opposite to this central angle is (a) 10ft (b) 3.6ft (c) 37.7ft.
16) The radius of a circle is 2.5m. The arc length opposite to 1rd of central angle in this circle is (a) 3.14m (b) 2.5m (c) 53.7^{o}. click here.
17) A driveway in a culdesac of radius 60.0ft shares 15ft of edge with that culdesac. A kid draws two lines from the center of the culdesac to the ends of that edge. The angle between these two lines is (a) 0.25 radians (b) 15π (c) 0.25ft.
18) The formula that relates arclength S to central angle θ is (a) S = θ/R (b) S = Rθ (c) S = θ/2.
19) The formula that relates linear speed v to angular speed ω is v = Rω. (a) True (b) False click here.
Problem: A piece of paper is stuck to the outer edge of a bicycle wheel that spins at 180 rpm. The radius of the wheel is 1.00ft.
20) The frequency of the rotation of the paper piece is (a) 18Hz (b) 3.0Hz (c) 0.50Hz.
21) The angular speed of the paper piece is (a) 25.1 rd/s (b) 18.8 rd/s (c) 18Hz. click here.
22) The linear speed of the paper piece as it travels along the edge of the wheel is (a) 18.8 ft/s (b) 25.1 ft/s (c) 36 ft/min.
23) The linear distance that the paper piece travels in one second is (a) 3 circumferences (b) 6 circumferences (c) 0.
24) Each circumference is (a) πR^{2} (b) πR (c) 2πR.
25) Since the linear speed of the paper piece is 3 circumferences per second, it can be written as V = 3(2πR )/s. This is equivalent to writing V=R 2π(3/s) or, V = R2πf, or V=Rω. (a) True (b) False
(26) Since f = 1/T, the formula V = R2πf in Question 25 can be written as V = 2πR/T. (a) True (b) False
The Velocity Vector:
Speed in circular motion can be kept constant. Velocity is never constant in circular motion because its direction keeps changing. Speed is constant only for uniform circular motion of a particle. As an example, when you turn a ceiling fan on, for a while, it is speeding up and therefore accelerating. During the acceleration phase, the motion is NOT uniform. When the fan reaches its maximum rpm, then each particle (or point) of it will have its own constant speed along its own circular path.
The Acceleration Vector:
In circular motion, there are two types of acceleration : "tangential acceleration" and "centripetal acceleration."
Tangential acceleration (a_{t}) is the result of the change in the speed of the object. In uniform circular motion where speed is constant, tangential acceleration is zero.
Centripetal acceleration (a_{c}) is the result of the change in the direction of the velocity vector of a rotating object. This acceleration is never zero because, in circular motion, the change in the velocity direction is an ongoing process. The following figure shows two wheels: one spinning at a constant rpm, and one at an increasing rpm. For the left wheel a_{t }= 0, and for the right wheel, a_{t }≠ 0. For both wheels a_{c} ≠ 0. Whether the wheel is turning at constant rpm or not, the centripetal acceleration is never zero.
The Centripetal Acceleration:
The formula for centripetal acceleration is a_{c} = v^{2}/R. As was mentioned above, a_{c} is the result of change in the direction of the velocity vector with time. We will not go through the proof of this formula and just solve a couple of examples on it. We will then review the centripetal force again.
Example 10: A tiny rock is caught in the treads of a car tire spinning at 420rpm on a wheel balancing machine. The radius of the tire is 0.35m. Calculate the centripetal acceleration given to the rock in forcing it to travel on circular path.
Solution: a_{c} = v^{2}/R. v must be calculated first. v = Rω ; ω = 2πf ; f is revs per second.
f = 420 (rev/min) = 420(rev/60s) = 7.0 rev/s.
ω = 2π f ; ω = 2π (7.0 rev/s) = 44 rd/s.
v = Rω ; v = 0.35m(44 rd/s) = 15.4 m/s.
a_{c} = v^{2}/R ; a_{c} = [15.4m/s]^{2 }/0.35m = 680 m/s^{2}.
Example 11: A car has a centripetal acceleration of 2.3 m/s^{2} as it travels along a curved portion of a road. The radius of curvature of the portion is 250m. Knowing that the linear speed of the car is constant, determine (a) its linear speed v, (b) its angular speed ω, (c) the angle θ it travels in 6.0s, and (d) the fraction of the circle it travels during this period. The top view of the circular motion of the car is shown.
Solution: (a) a_{c }= V^{2}/R ; a_{c}R = V^{2} ; V =(a_{c}R)^{1/2} ; V = [2.3(m/s^{2}) x 250m]^{1/2} = 24m/s. (b) V = Rω ; ω = v/R ; ω = (24m/s)/250m ω = 0.096 rd/s (rd is borrowed). (c) ω = Δθ/Δt ; Δθ = ωΔt ; Δθ = (0.096 rd/s)(6.0s) = 0.576rd. (d) Each circle is 6.28rd ; thus, fraction = 0.576rd /6.28rd = 0.092 = 9.2%.

Important: Pay attention to the direction of vector a_{c} = 2.3m/s^{2} that is drawn toward the center. 
The Centripetal Force and Acceleration:
The centripetal force formula is indeed Newton's Second Law (F=Ma) in which a is replaced by a_{c} = v^{2}/R. In doing so F will then become F_{c} , the centripetal force. We get:
F_{c} = Ma_{c} or,
A Good Link to Try: http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=21
The Motion of a Car Along a Curved Road
Roads that are engineered (banked) provide a tilt (bank) in any of their curved portions that protects vehicles from slipping off the road specially under icy conditions. The tilt in a curved portion allows vehicles to safely have a greater speed along that portion compared to an untilted or unbanked portion of the same radius of curvature. We are going to derive a formula for the maximum speed of the motion of a vehicle along a curved portion for two cases: (1) for an unbanked road, and (2) for a banked road. The maximum speed is the speed at which the vehicle is on the verge of slipping off the road.
In case (1), the vehicle relies completely on friction, and in case (2) it relies completely on the road's tilt and not on friction at all. The real situation is a combination of both. Case (1) is studied in the following example and the formula is actually derived in there.
Example 12: An 840kg car is negotiating a curve on an unbanked road where it has to totally rely on friction. The static coefficient of friction of the tires with the road is 0.73 and the radius of curvature is 110m. What maximum speed can the car have before slipping off the road? In the first figure the car is being viewed from its back as it negotiates a road that curves to the left. In the second figure the top view is shown.
The Motion of a Car Along an Unbanked Road:
From the above example, we have come up with the formula (*) for the motion of a car along a curved road that is not banked and the reliance is 100% on static friction:
v^{2} = μ_{s} R g.
As you may have noticed, the mass M of the car was given, but not used in the solution. It is interesting to note that the maximum speed does not depend on the mass of the vehicle, and for a fixed radius, it depends on μ_{s} only. Of course, no one should try to drive alongside a curve at maximum possible speed! There is no guarantee that the coefficient of static friction is the same everywhere along that curve.
Example 13: A car can negotiate an unbanked curve of radius 55m at a maximum speed of 66 km/h before the danger for slipping is felt. Determine the coefficient of friction that exists between its tires and the road.
Solution: v = 66 km/h = 66000m /3600s = 18 m/s. (Use horizontal fraction bars).
v^{2} = μ_{s}Rg ; μ_{s} =_{ }v^{2}/Rg ; μ_{s } =_{ }(18m/s)^{2} / [55m x 9.8m/s^{2}] = 0.60.
The Motion of a Car Along a Banked Road:
Under icy conditions, when a vehicle travels along a curved road of radius R that is banked, the tilt angle of the road is the important element that determines the maximum speed. Since the normal force, N , must be perpendicular to the road, it is not parallel to the weight force, w. In this case N > w if the car is in motion. In fact, N adjusts itself such that its vertical component F_{v} balances the vehicle's weight w out and its xcomponent provides the needed centripetal force F_{c} for the curved motion of the car. The following is the force diagram from which maximum possible speed can be calculated:
Example 14: A car can travel along a banked road on a curved portion of radius 125m at a maximum speed of 45 km/h without relying on friction. Determine the tilt angle of that portion.
Solution: V = 45km/h = 45000m/3600s = 12.5 m/s ; the appropriate formula is:
tan θ = V^{2}/Rg = 12.5^{2} /(125x9.8) = 0.128 ; θ = 7.3^{o}.
Chapter 5 Test Yourself 3:
1) In uniform circular motion, the linear speed is constant; therefore, the linear velocity is also constant. (a) True (b) False click here.
2) In uniform circular motion, the velocity vector is always tangent to (a) the radius (b) the center (c) the circle.
3) In this chapter, the two formulas written as s = Rθ and v = Rω are (a) both wrong (b) one correct, one wrong (c) both correct.
4) Tangential acceleration is (a) the change in angular speed per second (b) the change in the linear speed per second as an object speeds up along a circular path (c) the change in radius per second. click here.
5) Centripetal acceleration is (a) due to change in the direction of velocity vector that keeps changing in circular motion (b) because of change in radius (c) none of a or b.
6) At any point on a circular path, the tangential and centripetal acceleration vectors are (a) parallel to each other (b) perpendicular to each other (c) normal to each other (d) b and c are the same and correct.
7) Centripetal acceleration vector that is always perpendicular to the path of motion and always directed toward the center of rotation, may also be called the normal acceleration. (a) True (b) False
8) Centripetal acceleration is proportional to (a) the speed, V (b) the square of speed, V^{2} (c) the radius. click here.
9) Centripetal acceleration is proportional to (a) the radius, R (b) the reciprocal of radius, 1/R (c) square of radius, R^{2}.
10) Overall, the formula that calculates the magnitude of the centripetal acceleration is (a) a_{c}=V^{2}R (b) a_{c}=VR^{2} (c) a_{c}=V^{2}/R .
11) Since F = Ma ; therefore, (a) F_{c} = MV^{2}/R (b) F_{c }= MV/R (c) F_{c} = MVR .
12) The speed, V, in formula F_{c} = MV^{2}/R can be found from v = Rω. (a) True (b) False click here.
13) It is reasonable to think that when driving along an unbanked curved road, the danger of slipping off the road is proportional to (a) speed (b) the square of speed (c) the reciprocal of speed (1/V) because the faster we drive, the lesser the danger.
14) The formula that relates the coefficient of static friction to other factors for motion along an unbanked road where we completely rely on friction is (a) V = μ_{s} (b) V =μ_{s}Rg (c) V^{2} = μ_{s}Rg.
15) The reason for the use of μ_{s}, the coefficient of static friction, in the formula V^{2} = μ_{s}Rg is that slipping should not occur. (a) True (b) False
16) For an engineered (banked) road, the greater the angle of tilt, (a) the lower the safe speed (b) the higher the safe speed (c) the smaller the coefficient of static friction. click here.
17) For a banked curved road, the tilt angle is proportional to (a) the square of speed (b) just speed (c) the reciprocal of speed (1/V).
18) The tilt angle, θ of a banked road is (a) directly proportional to g (b) inversely proportional to g (c) is directly proportional to 1/g (d) both b & c.
19) The tilt angle, θ of a banked road is (a) directly proportional to R (b) inversely proportional to R (c) is directly proportional to 1/R (d) both b & c.
20) In the formula tanθ = V^{2}/Rg, (a) frictional effect is taken into account (b) no reliance is made on friction (c) if we include friction, it makes the angle of tilt greater. click here.
Problems:
1) A car can negotiate a banked portion of a curved road at a maximum speed of 25 mph under icy conditions. The radius of curvature of the road is 64m. Calculate the road's tilt angle.
2) The maximum speed that a car can make it through a curved portion of a flat (nonengineered) road is 46km/h. The radius of curvature of the road is 42m. Find the coefficient of static friction between the road and its tires.
3) A small plastic pail of water is attached to a string and spun in a vertical plane. What minimum speed is required at the top for the water not to spill out of the inverted pail? The radius of rotation of the water portion is 85cm.
4) A car travels along a road as shown. Both top and bottom of the hill have the same radius of curvature of 22m. (a) What is the maximum speed of the car for not to lose contact with the road at A? (b) If the car resumes the same maximum speed along the entire road, how heavy does an 83kg person feel as the car travels at B? (c) What is the normal weight of the person in N?


5) Calculate (a) the angular speed of the Earth about its own axis that passes through its poles, and (b) the linear speed of the people who live on the equator. The radius of the earth is 6280km.
6) A metal cylinder has a radius of 25.0cm. At what minimum rpm must it spin about its vertical axis such that a small 40.0gram block in contact with its vertical wall does not slide down? The coefficient of static friction between the block and the wall of the cylinder is 0.650. Let g =9.81m/s^{2}.



The condition to be met for the block not to fall is F_{s} = w. 
Answers: 1) 11^{o} 2) 0.40 3) 2.9m/s 4) 33mph, 1620N, 810N
5) 7.27x10^{5} rd/s, 456 m/s 6) 74.2 rpm
Footnote:
(*) In Δs = RΔθ when both sides are divided by Δt, we only divided one of the factors on the right by Δt to get ΔS/Δt = R (Δθ/Δt) that would become v = Rω. See the following examples:
We know that 48 = 12 + 36. This equality has 1 term on its left side and 2 terms on its right. Quantities separated by a + or a  sign are considered separate "terms." When an equality is divided through by a quantity, each term must be divided by that quantity. For example, if both sides of 48 = 12 + 36 are divided by 4, we get 12 = 3 + 9 that is a new equality but correct.
Now look at 48 = (4)(12) that is another correct equality. This one has 1 term on its left and 1 term on its right side. (4)(12) has two factors in it but is only 1 term. If we divide both sides of it by 4, we write the result as 12 = (1)(12) or 12 = (4)(3). Only one of the factors must be divided by 4 not both. As you see both new equalities are correct.
A number of quantities in the form of multiplication are factors to each other but we treat them as a single term. When such quantity is divided by a number, only one of its factors must be divided by that number.
It would be wrong to divide both sides of 48 = (4)(12) by 4 and write it as 12=(1)(3) !!! The equality does not hold true! The sides are not equal.