Chapter 5

Uniform Circular Motion

Younes Sina

Radian, the SI unit of Angle.One radian of angle is the central angle in any circle which opposite arc is equal tothe radius of that circle.

Example : Naming 3.14 rd as " π ", calculate angles360o, 180o, 90o, 60o, 45o, and 30o in terms of π.

Fraction: 1 full circle 1/2 circle 1/4 circle 1/6 circle 1/8 circle 1/12 circle

Degrees: 360o 180o 90o 60o 45o 30o

Radians: 2π = 6.28rd π π/2 π/3 π/4 π/6

R

D

2

360

180

.DR

Arc Length-Central 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 isvalid only if the central angle is measured or expressed in radians.

Example : Referring to the following figure, suppose angleθ is 148o and R = 1.25 in. Calculate the length of arc AB.

A

B

Solution:Using S = R θ and converting degrees to radiansyields: S = (1.25 in.)(148o)( 3.14rd / 180o ) = 3.23 in.

Angular Speed ( ω ):Angular speed ( ω ) is defined as the change in angle per unit of time. Mathematically, it may be written asω = Δθ / Δt .The preferred unit for angular speed is rd /s.The commercial unit is (rpm) or revolutions per minute.Note that each revolution is 6.28=(2π) radians.

Example : A car tire spins at 240 rpm.Calculate (a) its angular speed in rd /s(b) the angle that any of its radii sweeps in 44 seconds, and (c) the arc-length that any point on its outer edge travels during this timeknowing that R = 14 in.

Solution: 240 rpm is the angular speed (ω).(a) All we need to do is to convert it from rpm to rd/s

Note: 1 rev = 6.28 rd1min = 60 sec

ω = 240 (rev / min) = 240 ( 6.28 rd / 60sec) = 25 rd/sec

(b) ω = Δθ / ΔtΔθ = ω. ΔtΔθ = ( 25 rd/sec)(44 sec)Δθ = 1100 rd

(c) s = Rθs = ( 14 in)(1100 rd) = 15400 in = 1283 ft = 0.24 miles

Linear Speed - Angular Speed Formula:The same way S and θ are related, we have a formula that relates v to ω.v = Rω Writing these two similar formulas together helps their recall.s = Rθ and v = Rω

Derivation:All you need to do is to divide both sides of s = Rθ by tS/t = R θ /tv=S/t [distance over time]ω =(θ /t ) is ω, angle swept over timev = Rω

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

ω=25 rd/sec

Solution:

(a) v = Rωv = (14 in)(25 rd/sec) = 350 in/sec(b) s = vts = (350 in/sec)(44 sec) = 15400 inNote: In (b) equation x = (1/2)at2 + vi t is used wherea = 0 and x is replaced by s.This is because the arc length on 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 equationx = V t or s = V t is valid.

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.“

T = 1/ff = 1/T

Example : 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:

T = (1/2)s

f=2

If the frequency (f) is 3 per second, the time of each turn or (T) is T = (1/3)s.

linear speed (v) in a circular motion

T

R

t

xv

2

Example : A car tire is spinning at a constant rateof 660 rpm when the car's linear speed is 22 m/salong the road. Determine(a) the frequency of rotation of the tire(b) the period of rotation(c) the radius of the tireNote: π = 3.14

T

R

t

xv

2

Example : The Earth goes around the Sun once peryear (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.Suppose the orbit is a perfect circle although it is not.

Solution:R=150,000,000 kmT= 1 yearUsing the above formula:V = ( 2πR) / TV= ( 2π)(150,000,000km) / (365 x 24 x 3600s)V= 30 km/s = 19 mi/s

T

R

t

xv

2

The Relation Between ( f ) and ( ω ):

ω = 2π f

f is the # of turns per secondω the # of radians per second

( 1turn = 2π radians)

Example : Solve the previous problem usingω = 2π f and v = Rω.

Solution:

ω = 2π fω = 2π (1turn / year)ω = 2π /yearv = Rωv = ( 150,000,000km) ( 2π /year )= 30 km/s

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 andtherefore 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.

Acceleration Vector:In circular motion, there are two types of acceleration :"tangential acceleration" and "centripetal acceleration.“Tangential acceleration ( at ) 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 (ac ) 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 an increasing rpm. For the left wheel at = 0, and for the right wheel, at ≠ 0. However, for both wheels ac ≠ 0. Whether the wheel is turning at constant rpm or not, the centripetal acceleration is never zero.

centripetal acceleration:ac = v2/R

Example: A tiny rock is caught in the treads of a car tirespinning at 420 rpm on a wheel balancing machine.The radius of the tire is 0.35m.Calculate the centripetal acceleration given to the rock inforcing it to travel circularly.

Solution:

f means rev per secondf = 420 (rev / min)= 420(rev / 60s)= 7.0 rev/sω = 2π fω = 2π ( 7.0 rev / s) = 44 rd /sv = Rωv = (0.35m)(44 rd/s) = 15.4 m/sac = v2 / Rac = [15.4(m/s)]2 / 0.35m = 680 m/s2

Example : 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 250 m.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 6.0 s(d) the fraction of the circle it travels during this period.The top view of the circular motion of the car is shown.

V=?

ω=?

Δθ =?

Δs=?

Solution:

(a) ac = v2/RacR = v2

v =(acR)1/2

v = [2.3(m/s2) x 250m]1/2 = 24m/s(b) v = Rω v/R = ωω = 24(m/s)/250mω = 0.096 rd/s(c) ω = Δθ/ΔtΔθ = ωΔtΔθ = (0.096 rd/s)(6.0s) = 0.576 rd(d) Each circle is 2π = 6.28 rdfraction = 0.576 rd /6.28 rd = 0.092 = 9.2%

V=?

ω=?

Δθ =?

Δs=?

Centripetal Force: Objects tend to move along straight lines unless they areforced to do otherwise. When an object is moving along a circle, it is forced to have that circular path. The force pulls or pushes the object toward the center of rotation. That is why it is called the "centripetal force, Fc." 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 (via the string) 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 the centripetal force, the force that pulls the object toward the center of rotation, no matter where the object is in its circular path. An object of mass M that has a centripetal acceleration, ac, along its circular path is therefore under a force of Fc = Mac. The formula for centripetal force becomes:

Fc = Mv2 / Rcentripetal force

Example : A rock of mass 0.22 kg is attached to a string of length 0.43 m and is given a circular motion in a horizontal plane at a rate of 180 rpm.Calculate the centripetal force that the string exerts on the rock pulling it nonstop toward the center of rotation. Draw a circle and show M, V, R , and Fc on it as shown above.

Solution:ω = 180 rev / min = 180( 6.28 rd / 60s) = 18.84 rd/sv = Rωv = (0.43m)(18.84 rd/s) = 8.1 m/sFc = Mv2 / RFc = 0.22kg (8.1 m/s)2 / 0.43m = 34N

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 un-tilted or un-banked portion of the same radius of curvature.

In case of unbanked road, the vehicle relies completely on friction.

W=N= M.g

a= v2/R

Fc =Fs NR s

MV 2

MgR s

MV 2

gR s

V 2

V2=μsgR

In case of a banked road the vehicle relies completely on the road's tilt and not on friction at all.

tan (θ)=Fc/W

RgMgRMg

R VMVMV 22

2

)tan(

V2=Rg.tan(θ)

Example : A car can negotiate an un-banked 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.

V2=μsgRR=55 m V=66 km/h μs=?

Solution:v = 66 km/h ( 1000m / km )( 1h / 3600s )= 18.3 m/s v2 = μs R gμs = v2 / R gμs = (18.3m/s)

2/ (55m X 9.8m/s

2) = 0.62

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

R=125 m V=45 km/h V2=Rg.tan(θ)

Solution:V = 45km/h = 45000m / 3600s =12.5 m/s

tan θ = V2 / Rgtan θ = 12.52 / (125x9.8) = 0.128θ = 7.3o

Homework : problems 1, 2, 3, and 5 chapter 5