PHYS101 Uniform Circular Motionopencourses.emu.edu.tr/.../content/1/uniform_circular_motion.pdf ·...

5
PHYS101 Uniform Circular Motion Uniform Circular Motion Uniform circular motion means, that the particle moves with constant speed on a circle. So, the first question that arises is: Is the particle accelerated? Recall: ~ a(t)= d ~ v(t) dt In order to answer this question, recall that the velocity of a particle is a vector prop- erty, i.e. it has a magnitude and a direction. As uniform circular motion means that the speed is constant, i.e. the magnitude of the velocity, there is no restriction to the direction. Obviously, the direction is of the velocity is changing. Figure 1: Particle moving in uniform circular motion Let the motion be counter clockwise so, that the angle ϕ(t) > 0. Then we can write for the position vector: ~ r(t)= R cos ϕ(t) ˆ i + R sin ϕ(t) ˆ j, (1) where R is the constant radius of the circle and ϕ(t) is the angle that changes over time, changing also the direction of the position vector. Now, let us calculate the velocity, as the sec- ond requirement was that the speed of the particle on the circle is constant. ~ v(t)= ~ r(t) dt = -R sin ϕ(t) d ϕ(t) dt ˆ i + R cos ϕ(t) d ϕ(t) dt ˆ j = = R d ϕ(t) dt ( - sin ϕ(t) ˆ i + sin ϕ(t) ˆ j ) (2) So, we can now calculate the speed, which is the magnitude of the velocity ~ v(t) as: | ~ v(t)| = s -R sin ϕ(t) d ϕ(t) dt 2 + R cos ϕ(t) d ϕ(t) dt 2 = s R 2 sin 2 ϕ(t) d ϕ(t) dt 2 + R 2 cos 2 ϕ(t) d ϕ(t) dt 2 = v u u u t R 2 d ϕ(t) dt 2 sin 2 ϕ(t)+ cos 2 ϕ(t) 2 | {z } =1 = s R 2 d ϕ(t) dt 2 = R d ϕ(t) dt = const. (3) c 2015 Department of Physics, Eastern Mediterranean University Page 1 of 5

Transcript of PHYS101 Uniform Circular Motionopencourses.emu.edu.tr/.../content/1/uniform_circular_motion.pdf ·...

Page 1: PHYS101 Uniform Circular Motionopencourses.emu.edu.tr/.../content/1/uniform_circular_motion.pdf · PHYS101 Uniform Circular Motion Tangential and Centripetal Acceleration Figure 3:

PHYS101 Uniform Circular Motion

Uniform Circular Motion

Uniform circular motion means, that the particle moves with constant speed on acircle.So, the first question that arises is: Is the particle accelerated?Recall:

~a(t) =d~v(t)

dtIn order to answer this question, recall that the velocity of a particle is a vector prop-erty, i.e. it has a magnitude and a direction. As uniform circular motion means thatthe speed is constant, i.e. the magnitude of the velocity, there is no restriction to thedirection. Obviously, the direction is of the velocity is changing.

Figure 1: Particle moving in uniformcircular motion

Let the motion be counter clockwise so, thatthe angle ϕ(t) > 0. Then we can write for theposition vector:

~r(t) = R cos ϕ(t) i + R sin ϕ(t) j, (1)

where R is the constant radius of the circleand ϕ(t) is the angle that changes over time,changing also the direction of the positionvector.Now, let us calculate the velocity, as the sec-ond requirement was that the speed of theparticle on the circle is constant.

~v(t) =~r(t)dt

= −R sin ϕ(t)dϕ(t)

dti+R cos ϕ(t)

dϕ(t)dt

j =

= Rdϕ(t)

dt(− sin ϕ(t) i + sin ϕ(t) j

)(2)

So, we can now calculate the speed, which is the magnitude of the velocity~v(t) as:

|~v(t)| =

√(−R sin ϕ(t)

dϕ(t)dt

)2

+

(R cos ϕ(t)

dϕ(t)dt

)2

=

√R2 sin2 ϕ(t)

(dϕ(t)

dt

)2

+ R2 cos2 ϕ(t)(

dϕ(t)dt

)2

=

√√√√√R2(

dϕ(t)dt

)2 (sin2 ϕ(t) + cos2 ϕ(t)

)2

︸ ︷︷ ︸=1

=

√R2(

dϕ(t)dt

)2

= Rdϕ(t)

dt= const. (3)

c©2015 Department of Physics, Eastern Mediterranean University Page 1 of 5

Page 2: PHYS101 Uniform Circular Motionopencourses.emu.edu.tr/.../content/1/uniform_circular_motion.pdf · PHYS101 Uniform Circular Motion Tangential and Centripetal Acceleration Figure 3:

PHYS101 Uniform Circular Motion

As the radius of the circle R is constant, and the speed of the particle must be constantas well, then the change of the angle with respect to time should be constant as well

dϕ(t)dt

= const. (4)

The solution of the so-called differential equation (4) can be obtained very straightforward as:

ϕ(t) = ωt + ϕ0 (5)

where ω and ϕ0 are constant. We will identify ω later as the angular speed, and ϕ0 asthe initial angle. The angular speed is in the case of uniform circular motion constant.We can check now, if (5) is the solution of (4), by differentiating ϕ(t) with respect to t

ddt

ϕ(t) =ddt

(ωt + ϕ0) = ω = const.

So, we can accept (5) as the solution to (4). So, let us now rewrite the position vector,the velocity vector, and acceleration vector. So the position vector becomes:

~r(t) = R cos (ωt + ϕ0) i + R sin (ωt + ϕ0) j. (6)

The magnitude and the direction are then:

|~r(t)| = R, θ = ϕ(t) = ωt + ϕ0

The velocity vector is

~v(t) = −Rω sin (ωt + ϕ0) i + Rω cos (ωt + ϕ0) j, (7)

where the magnitude and direction of the velocity vector are:

|~r(t)| = Rω, θ = ϕ(t) + 90◦ = ωt + ϕ0 + 90◦.

Using the identitycot θ = tan(90◦ − θ)

we can easily verify the direction as following:

θ = tan−1(

cos(ωt + ϕ0)

− sin(ωt + ϕ0)

)= tan−1 (− cot(ωt + ϕ0)) =

= tan−1(

tan(90◦ − (−(ωt + ϕ0))

))= 90◦ + (ωt + ϕ0)

Finally, let us consider the acceleration vector:

~a(t) =d~v(t)

dt= −Rω2 cos (ωt + ϕ0) i− Rω2 sin (ωt + ϕ0) j =

= −ω2 (R cos (ωt + ϕ0) i + R sin (ωt + ϕ0) j)= −ω2~r(t) (8)

c©2015 Department of Physics, Eastern Mediterranean University Page 2 of 5

Page 3: PHYS101 Uniform Circular Motionopencourses.emu.edu.tr/.../content/1/uniform_circular_motion.pdf · PHYS101 Uniform Circular Motion Tangential and Centripetal Acceleration Figure 3:

PHYS101 Uniform Circular Motion

Figure 2: Graphical illustration of thefound results

Equation (8) obviously shows that the direc-tion of the acceleration vector is in oppositedirection to the position vector, i.e. the accel-eration vector points towards the center of thecircle, and is therefore called centripetal accel-eration. The word centripetal means towardsthe center, or center seeking. Then we can cal-culate the magnitude and direction of the ac-celeration vector:

ac = |~a(t)| = ω2 |~r(t)| = ω2R =

=ω2R2

R=

v2

RSummarising, the magnitude of the cen-tripetal acceleration~ac

ac =v2

R(9)

The direction of the acceleration vector can be easily determined as

θ = 180◦ + ωt + ϕ0.

As the speed is constant, the average speed and the instantaneous speed are equal.The average speed can be calculated as

v =d

∆t,

where d is the distance travelled and ∆t is the time need to travel over the distanced. So we can easily determine the time T needed for one revolution, i.e. the time forone complete turn, as the particle travels over the distance d = 2πR with the constantspeed v. So, the time for one revolutions is:

T =2πR

v(10)

Example 1. What is the centripetal acceleration of the Earth as it moves in its orbitaround the sun?For the solution of this problem, we assume that the Earth is moving on a circlearound the sun, and that the speed of the Earth is constant on its orbit. We know:

T = 1a = 365.25d× 24hd× 60

minh× 60

smin

= 3.154× 107s, R = 1.496× 1011m

So, we get for the cetripetal acceleration:

ac =v2

R=

(2πRT)2

R=

4π2RT2 =

4π2 1.496× 1011m

(3.154× 107s)2 = 5.937× 10−3 ms2

c©2015 Department of Physics, Eastern Mediterranean University Page 3 of 5

Page 4: PHYS101 Uniform Circular Motionopencourses.emu.edu.tr/.../content/1/uniform_circular_motion.pdf · PHYS101 Uniform Circular Motion Tangential and Centripetal Acceleration Figure 3:

PHYS101 Uniform Circular Motion

Tangential and Centripetal Acceleration

Figure 3: Motion of a particle along anarbitrary curved line

We can also describe the motion of a particlealong an arbitrary curved line. As we havediscussed in the previous section, the cen-tripetal acceleration is only due to the changeof the direction of the velocity vector. But, wehave not discussed the change of the changeof the magnitude of the velocity, which causesan increasing speed if the acceleration vectorand the velocity vector point into the samedirection, or a deceleration if the accelerationvector points into the opposite direction of thevelocity vector. Considering the motion along

an arbitrary curved line we can inscribe at every point a circle and determine the cen-tripetal and tangential acceleration. The total acceleration is given as:

~a =~ac +~at , (11)

where the tangential acceleration causes a change in the speed and its magnitude is

at =

∣∣∣∣dvdt

∣∣∣∣ = ddt|~v(t)| . (12)

As one can see easily from figure 3, it is obvious that the tangential and centripetalaccelerations are perpendicular to each other.

Example 2. A car leaves a stop sign and exhibits a constant acceleration of 0.3m/s2

parallel to the roadway. The car passes over a rise in the roadway such that the top ofthe rise is shaped like an arc of a circle of radius 500m. At the moment the car is at thetop of the rise, its velocity vector is horizontal and has a magnitude of 6m/s. What arethe magnitude and direction of the total acceleration vector for the car at this instant?

Figure 4: A car passes over a rise, shaped like a circle

The magnitude of the centripetal acceleration can be easily determined as:

ac =v2

R=

(6m/s)2

500m= 0.072

ms2

c©2015 Department of Physics, Eastern Mediterranean University Page 4 of 5

Page 5: PHYS101 Uniform Circular Motionopencourses.emu.edu.tr/.../content/1/uniform_circular_motion.pdf · PHYS101 Uniform Circular Motion Tangential and Centripetal Acceleration Figure 3:

PHYS101 Uniform Circular Motion

The total acceleration is the vector sum of the centripetal acceleration and the tangen-tial acceleration.

~a =~at +~ac = 0.3ms2 i− 0.072

ms2 j

Figure 5: Accelerations acting on the car

Then, the magnitude of the acceleration is

a =√

a2t + a2

c =

√(0.3

ms2

)2+(−0.072

ms2

)2= 0.309

ms2 .

The angle φ with the positive x-axis is then:

φ = tan−1(

ac

at

)= tan−1

(−0.072 m

s2

0.3 ms2

)= −13.5◦

Example 3 (Moon’s centripetal acceleration). The Moon’s nearly circular orbit aboutthe Earth has a radius of about 384000km and a period T of 27.3 days. Determine theacceleration of the Moon toward the Earth.The distance the Moon travels in 27.3 days is

d = 2πR = 2π3.84× 108m = 2.41× 109m.

The time for one revolution of the Moon about the Earth is

T = 27.3d = 27.3d× 24hd× 60

minh× 60

smin

= 2.36× 106s

So the average speed of the Moon around the Earth:

vavg =dT=

2.41× 109m2.36× 106s

= 1.0211× 103 ms

So, finally we get for the centripetal acceleration

ac =v2

R=

(1.0211× 103 m

s)2

2.41× 109m= 0.266× 10−3 m

s= 2.78× 10−3 m

s2 .

c©2015 Department of Physics, Eastern Mediterranean University Page 5 of 5