Chapter 5

One radian of angle is the central angle in any circle whose opposite arc is equal to the radius of that circle. If a piece of string is cut equal to the radius 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."

Example 1: Naming 3.14rd as " π ", calculate angles 360°, 180°, 90°, 60°, 45°, and 30° in terms of π.

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° 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°)( 3.14rd / 180° ) = 3.2 in.

This symbol ( ω ) is the lower case of symbol ( 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 . A preferred unit for angular speed is ( rd / s ). A commercial unit is (rpm) or revolutions per minute.

Example 3: A car tire spins at 240rpm. Calculate (a) its angular speed in the preferred unit of (rd /s), (b) the angle that a typical radial line on it (such as a spoke) sweeps in 44 seconds, and (c) the arc length that a point on its outer edge travels during this time knowing that R = 14 in.. Make sure that you completely sole this problem on paper using horizontal fraction bars everywhere.

Solution: 240rpm is ω itself, the angular speed. 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 / 60sec) = 25 rd/sec. Note: 1rev = 6.28 rd, and 1min = 60 sec..

(b) ω = Δθ / Δt or Δθ = ωΔt or Δθ = ( 25 rd/sec)(44 sec) or Δθ = 1100 rd.

The same way S and θ are related, we have a formula that relates v to ω. The formula is: v = Rω .

Writing these two similar formulas together helps their recall. s = Rθ and v = Rω.

The derivation is easy. All you need to do is to divide both sides of s = Rθ by (t) to get v = Rω.

(s/t) = R (θ /t) (s / t) is v, distance over time. ( θ /t ) is ω, angle over time; therefore, v = Rω.

Example 4: The radius of a car tire is 14 in. Calculate (a) the linear speed (v) of a point on its outer edge, if it spins at an angular speed of 25 rd/sec. (b) Find the linear distance (arc length) the point travels in 44 sec..

(b) s = vt or s = (350 in./sec.)(44 sec.) = 15400 in. Use horiz. fraction bars when you solve.

Note: In (b) equation x = (1/2)at² + V_i t is used where a = 0 and x is replaced by s . This is because the arc length over the tire is like a long string wrapped around it and as the car moves, it leaves the unwrapped string on the ground as a straight line for which equation x = V t or s = V t is valid.

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 a central angle is equal to (a) the arc opposite to it when expresses 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° and one equal to 45°. Since 90° is 1/4 of 360°, verify that the arc-length opposite to the 90°-angle you chose is also 1/4 of the whole circle. Also, verify that the 45°-angle you chose is opposite to an arc that is exactly 1/8 of the whole circle. What conclusion do you draw? State ......................Is your conclusion in line with the correct answer to Question 3? click here

5) 1rd is the central angle whose opposite arclength equals (a) the radius of the circle. (b) the diagonal of the circle. (c) the perimeter of the circle.

6) The central angle whose opposite arc equals the perimeter of the circle is (a) 360°. (b) 2π radians. (c) both a and b.

7) The central angle whose opposite arc equals half of the circle is (a) 120°. (b) 180°. (c) 3.14radians. (d) both b and c.

8) 2.00 radians of angle is equivalent to (a) 114.6°. (b) 120°. (c) 170°. 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 whose radius is R = 6.00" is (a) 12.0". (b) 8.00". (c) 18.0".

11) The arc length opposite to a 114.6° central angle in a circle whose radius is 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 a 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 150. rd of angle every minute. Its angular speed is (a) 150.rd/s (b) 150.rd/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) a commercial unit of angular speed.

17) rd/s is (a) a commercial 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

21) 3600 rpm is equivalent to (a) 60 rev/sec. (b) 60x6.28rd / sec. (c) 377 rd/s (d) a, b, and c.

22) A helicopter propeller rotates at 956 rpm. It 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 at²

24) A helicopter propeller rotates at 956 rpm. The linear speed of the tip of its propeller that is 5.00m long is (a) 500.m/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) 500.m/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 regardless of their radius of rotation have the same (a) linear speed. (b) acceleration. (c) angular speed.

28) The linear speed of a point on a solid rotating disk (a) is a function of R, its distance from the center of rotation. (b) does not depend on R, its distance from the center. (c) is a function of the angular speed of the disk. (d) both a and c.

29) All points on the periphery 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.

When a particle of mass M moves along a circular path at 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.

Example 6: A car tire is spinning at a constant rate of 660rpm when the car's linear speed is 22m/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.

Example 7: The Earth goes around the Sun once per year (as we know). Calculate the linear speed of the Earth as it goes around the Sun. The average distance between the two is 150 million kilometers.

Solution: The orbit is so big that the almost circular path feels like moving along a straight line. The radius of rotation, R, is given, but in (km). The period (T) is 1 year. Using the above formula:

V = ( 2πR) / T = ( 2π)(150,000,000km) / (365 x 24 x 3600s) = 30 km/s = 19 mi/s

There is a useful formula that relates ω to f . Since f is the # of turns per second and ω the # of radians per second and we know that ( 1turn = 2π ), therefore, ω = 2π f .

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 (a) the time for 1 radian of angle traveled. (b) time for completing 100 turns. (c) time for completing one turn. click here

6) Frequency, f, is (a) the # of turns per second. (b) the # of rotations per second. (c) the reciprocal of period, T. (d) a, b, and c.

7) When f = 5s^-1, it means that the circulating 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/sec. (c) f = 5 Hertz (d) f = 5 Hz. (e) all of them.

9) f = 10 rev/s is the same thing as (a) f = 98.6 rd/s. (b) 31.4 rd/s. (c) f = 10s^-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 traveled per second) is ω=2πf. (a) True (b) False click here

We may write (a) f = 10s^-1 and ω = 62.8 /s (b) f = 10s^-1 and ω = 62.8 rd/s (c) f = 10s and ω = 62.8 rd/s.

(a) f = 8s^-1 and ω = 480 /s (b) f = 480s^-1 and ω = 8 rd/s (c) f = 8 / s and 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 cul de sac has a radius of 60.0 ft. Two straight lines from the center of the cul de sac are drawn to its circular edge. The angle in between measures 36º. The arc length opposite to this central angle is (a) 1 rd (b) 3.6rd (c) 37.7ft

16) The radius of a circle is 2.5m. The arc length opposite to 1 rd of central angle in this circle is (a) 3.14m (b) 2.5m (c) 53.7º click here

17) A driveway in a cul de sac of radius 60.0 ft shares 15ft of edge in that cul de sac. A kid draws two lines from the center of the cul de sac 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

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.

21) The angular speed of the paper piece is (a) 25.14 rd/s (b) 18.84 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) 18.84 ft/s (b) 25.14 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

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 = R 2π f, or V = R ω. (a) True (b) False

(26) Since f = 1/T, the formula V = R 2π f in Question 25 can be written as V = 2πR / T. (a) True (b) False

Speed in circular motion can be 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 circle of rotation.

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, the tangential acceleration is zero.

Centripetal acceleration (a_c) is the result of the change in the direction of velocity of the object. This acceleration is never zero, because the change in direction is an ongoing process, in circular motion. The following figure shows two wheels: one spinning at constant rpm, and one at increasing rpm. For the left wheel a_t= 0, and for the right wheel, a_t≠ 0. However, for both wheels a_c ≠ 0. Whether the wheel is turning at constant rpm or not, the centripetal acceleration is never zero.

The formula for centripetal acceleration is a_c = v²/R . At this point, let's just accept this formula and solve one numerical example on it, then introduce "centripetal force," and also solve a couple of numerical examples on it, and finally investigate the reason why centripetal force is directed toward the center of rotation.

Example 9: A tiny rock is caught in the treads of a car tire spinning at 420rpm. The radius of the tire is 0.35m. Calculate the centripetal acceleration given to the rock.

Solution: a_c = v²/ R. (v) must be calculated first. v = Rω ; ω = 2π f ; (f) means rev per second.

Example 10: A car has a centripetal acceleration of 2.3 m/s² as it travels along a curved portion of a road. The road has a radius of curvature of 250m. Knowing that the linear speed of the car is constant, determine (a) its linear speed, (b) its angular speed, (c) the angle it travels in 6.0s, and (d) the fraction of the circle it travels during this period.

Solution: (a) a_c = v²/ R ; a_cR = v² ; v = ( a_cR ) ^1/2 ; v = (2.3x250)^1/2 = 24m/s

(d) Each circle is 6.28rd ; therefore, fraction = 0.576rd / 6.28rd = 0.092 = 9.2%

Centripetal Force: Objects tend to move along straight lines unless they are forced to do otherwise. When an object is moving along a circle, it is forced to have a circular path. An object of mass M that has a centripetal acceleration, a_c, along its circular path is therefore under a force of F_c = Ma_c. The formula for centripetal force becomes:

Example 11: A rock of mass 0.22kg is attached to a string of length 0.43m and is given circular motion in a horizontal plane at a rate of 180rpm. Calculate the centripetal force that the string exerts on the rock pulling it nonstop toward the center of rotation. Draw a circle and show a velocity vector V, R( the radius that goes from the center of the circle to the tail of the velocity vector), and F_c (that goes from the tail of V toward the center) on it.

Solution: ω = 180 rev / min = 180( 6.28 rd / 60s) = 18.84 rd/s (Not rounded)

Example 12: An 840-kg 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 for not slipping off the road? In the first figure the car is being viewed from the back. In the second figure a top view is chown.

From the above example, we have come up with a general formula (*) for the motion of a car along an unbanked curve, simply:

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 driving alongside a curve at maximum possible speed!

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 ( 1000m / km )( 1h / 3600s ) = 18.3 m/s (Use horizontal fraction bars).

v² = μ_s R g ; μ_{s
=} v² / R g ; μ_{s
= (18.3m/s)}²_{/ (55m X 9.8m/s}²_{) =}_0.62

When a vehicle travels along a banked curve of radius ( R ), 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. Following is the force diagram from which maximum possible speed can be calculated:

Example 14: A car can travel along a banked curve of radius 125m at a maximum speed of 45 km/h without relying on friction. Determine the angle of the tilt of the road along this curve.

Solution: V = 45km/h = 45000m / 3600s = 12.5 m/s ; The appropriate formula is:

1) In uniform circular motion, the linear speed is constant; therefore, the linear velocity is also constant. (a) True (b) False

2) In uniform circular motion, the velocity vector is always tangent to (a) the radius. (b) the center. (c) the circle.

3) The formulas s = Rθ and v = Rω are (a) both false (b) one true, one false (c) both true

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.

5) Centripetal acceleration is (a) due to changes in the direction of velocity vector that keeps changing in uniform 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 magnitude of velocity, V. (b) the square of velocity magnitude, V². (c) the radius

9) Centripetal acceleration is proportional to (a) the radius, R. (b) the reciprocal of radius, 1/R. (c) the radius squared, R².

10) Overall, the formula that calculates the magnitude of the centripetal acceleration is (a) a_c=V²R. (b) a_c=VR². (c) a_c=V²/R.

12) The velocity, V in the centripetal acceleration / force formula can still be found by the formula v = Rω. (a) True (b) False.

13) It is reasonable to think that when driving along an unbanked curve, the danger of slipping off the road is proportional to (a) the magnitude of velocity. (b) the square of the magnitude of velocity. (c) the reciprocal of velocity (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 curve where we completely rely on friction is (a) V = μ_s. (b) V = μ_sRg. (c) V² = μ_sRg.

15) The reason for the use of μ_s, the coefficient of static friction, in the formula V² = μ_sRg is that slipping should not occur. (a) True (b) False

16) For an engineered (banked) curve of a road, the greater the angle of tilt, (a) the slower the safe speed. (b) the faster the safe speed. (c) the smaller the coefficient of static friction.

17) For a banked curve, the tilt angle is proportional to (a) the square of the magnitude of velocity. (b) the magnitude of velocity. (c) the reciprocal of velocity (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 tan θ = V² / 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.

1) A car can negotiate a banked portion of a 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 (non-engineered) road is 46km/h. The radius of curvature of the road is 42m. Find the coefficient of 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 is the minimum speed 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 83-kg person feel as the car travels at B? (c) What is the normal weight of the person in Newtons?

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 if the earth is 6280km.

6) A metal cylinder has a radius of 25.0cm. At what minimum rpm must it be spinning about its vertical axis such that a small block of mass 40.0grams in contact with its vertical wall does not slide down? The coefficient of friction between the block and the wall of the cylinder is 0.650. Let g =9.81m/s².

The condition for not falling of the block is

F_s = w.

Fraction:	1 full circle	1/2 circle	1/4 circle	1/6 circle	1/8 circle	1/12 circle
Degrees:	360°	180°	90°	60°	45°	30°
Radians:	2π = 6.28rd	π	π/2	π/3	π/4	π/6