Circular Motion Kinematics 8.01 W04D1. Today’s Reading Assignment: W04D1 Young and Freedman: 3.4; 5.4-5.5 Supplementary Notes: Circular Motion Kinematics.

Circular MotionKinematics

8.01

W04D1

Today’s Reading Assignment: W04D1

Young and Freedman: 3.4; 5.4-5.5

Supplementary Notes: Circular Motion Kinematics

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Learning Objectives1. Calculate time derivative of vector of constant magnitude2. Recognize that circular motion is a constraint (like sliding

down an inclined plane)3. Calculate direction and magnitude of velocity and

acceleration for circular motion when:i) constant speedii) non-constant speed

4. Explain difference between polar and cartesian coordinates

5. Express position, velocity, and acceleration in polar coordinates

6. Explain meaning of angular velocity, angular frequency, and angular acceleration

7. Know how to use different expressions for radial component of acceleration in terms (v,ω) and then in terms of (r, v), (r,ω), (r, f), and (r,T)

Concept Question: Time Derivative of a Vector

The time derivative of a vector1. is always zero.2. is non-zero if the magnitude of the vector is changing in

time.3. is non-zero if the direction of the vector is changing in time.4. can be zero if the both it’s magnitude and direction are

changing in time in such a way as to add to zero.5. Two of the above6. None of the above.

Circular Motion Kinematics

Concept Question: Time Derivative of Position Vector for

Circular Motion

A point-like object undergoes circular motion at a constant speed. The vector from the center of the circle to the object

1. has constant magnitude and hence is constant in time.

2. has constant magnitude but is changing direction so is not constant in time.

3. is changing in magnitude and hence is not constant in time.

Position and Displacement

: position vector of an object moving in a circular orbit of radius R

: change in position between time t and time t+t

Position vector is changing in direction not in magnitude.

The magnitude of the displacement is the length of the chord of the circle:

( )trr

( )trr

2 sin( / 2)R θ = rr

Review: Definition of Instantaneous Velocity

At time t, consider a finite time interval , calculate the displacement . As , generate a sequence of average velocities. The limiting value of this sequence is defined to be the instantaneous velocity at the time t.

0t →( )trr t

Direction of Velocity

Sequence of chord directions approach direction of velocity as approaches zero.

The direction of velocity is perpendicular to the direction of the position and tangent to the circular orbit.

Direction of velocity is constantly changing.

rr

t

Small Angle ApproximationWhen the angle is small:

Power series expansion

Using the small angle approximation with ,

the magnitude of the displacement is

sinφ ≈φ, cosφ ≈1

sinφ =φ−φ3

3!+

φ5

5!−...

cosφ =1−φ2

2!+

φ4

4!−...

2 sin( / 2)R Rθ θ = ≈ rr

φ =θ / 2

Speed and Angular Speed

The speed of the object undergoing circular motion is proportional to the rate of change of the angle with time:

Angular speed:

0 0 0lim lim lim

t t t

R dv R R R

t t t dt

θ θ θ ω → → →

≡ = = = = =

r

vr

r

ω =

dθdt

(units:rad⋅s-1)

Circular Motion: Constant Speed, Period, and Frequency

In one period the object travels a distance equal to the circumference:

Period: the amount of time to complete one circular orbit of radius R

Frequency is the inverse of the period:

s =2πR=vT

2 2 2R RT

v R

π π πω ω

= = =

11(units: s or Hz)

2f

T

ωπ

−= =

Table Problem: Demo Rising Ping Pong Ball

A wheel is connected via a pulley to a motor. A thread is knotted and placed through a hole in a ping pong ball of mass m. A wing nut secures the thread holding the ball a distance R from the center of the wheel. The wheel is set in motion. When a satisfactory angular speed ω is reached, the string is cut and the ball comes off at a tangent to the spinning wheel, traveling vertically upward a distance h. Find the frequency of the rotating wheel just before the string is cut.

Demo: Centripetal vs. Centrifugal B102

1. http://tsgphysics.mit.edu/front/index.php?page=demo.php?letnum=B%20102&show=0

A wooden ball is attached to the rim of a spinning wheel. The ball is held in place by a string. When the string is cut, the ball flies in a straight tangent to the wheel.

Cylindrical Coordinate System

Coordinates

Unit vectors

(r,θ,z)

ˆˆ ˆ( , , )θr z

Circular Motion: Vector Description

Use plane polar coordinates

Position

Velocity

ˆ( ) ( )t R t=r rr

rv(t) =R

dθdt

θ̂(t) =Rω θ̂(t)

Acceleration and Circular Motion

When an object moves in a circular orbit, the direction of the velocity changes and the speed may change as well.

For circular motion, the acceleration will always have a radial component (ar) due to the change in direction of velocity.

The acceleration may have a tangential component if the speed changes (at). When at =0, the speed of the object remains constant.

Direction of Radial Acceleration: Uniform Circular

Motion

Sequence of chord directions approaches radial inward direction as approaches zero

Perpendicular to the velocity vector

Points radially inward

vr

t

Table Problem: Magnitude of Change in Velocity for Circular

MotionFind the change in the magnitude of the velocity in terms of v and Δθ. Use the small angle approximation as necessary.

Magnitude of Change in Velocity: Circular Motion

SolutionChange in velocity:

Magnitude of change in velocity:

Using small angle approximation

( ) ( )t t t = + −v v vr r r

2 sin ( / 2)v θ = vr

v θ ≅ vr

Radial Acceleration: Constant Speed Circular

Motion Any object traveling in a circular orbit with a constant speed is always accelerating towards the center.

Direction of velocity is constantly changing.

Radial component of ar (minus sign indicates direction of acceleration points towards center)

ra

r≡arr̂ = lim

t→ 0

rv

t=−lim

t→ 0

vθt

r̂ =−vlimt→ 0

θt

r̂ =−vdθdt

r̂ =−vω r̂ =−v2

Rr̂

a

r=−

v2

R=−ω 2R

Concept QuestionA car is rounding a circular turn of radius R=200m at constant speed. The magnitude of its centripetal acceleration is 2 m/s2. What is the speed of the car?

1. 400 m/s2. 20 m/s3. 100 m/s4. 10 m/s5. None of the above.

Alternative Forms of Magnitude of

Radial Acceleration

Parameters: speed v, angular speed ω, angular frequency f, period T

2 22 2

2

4(2 )r

v Ra R R f

R T

πω π= = = =

Concept Question: Circular Motion

As the object speeds up along the circular path in a counterclockwise direction shown below, its acceleration points:

1. toward the center of the circular path.

2. in a direction tangential to the circular path.

3. outward.

4. none of the above.

Tangential Acceleration Circular Motion: Magnitude of Velocity is

Non-ConstantWhen the magnitude of the velocitychanges by an amount Then the tangential acceleration is the time rate of change of the

magnitude of the velocity

Angular acceleration: rate of change of angular velocity with time

ra

t=atθ̂ = lim

t→ 0

vt

θ̂ =Rlimt→ 0

ωt

θ̂ =Rdωdt

θ̂ =Rd2θdt2

θ̂ =Rα θ̂

α =

dωdt

=d2θdt2 (units:rad⋅s-2 )

v =Rω v=Rω

Circular Motion: Vector DescriptionUse plane polar coordinates

Position

Angular Speed

Velocity

Angular Acceleration

Acceleration

rr(t) =Rr̂(t)

rv(t) =R(dθ / dt) θ̂ =Rω θ̂

ra =ar r̂ + aθ θ̂

a

r=−rω 2 =−(v2 / r), aθ =rα

ω ≡dθ / dt

α ≡dω / dt =d2θ / dt2

rv =vθ θ̂(t)

Concept Question: Cart in a Turn

A golf cart moves around a circular path on a level surface with decreasing speed. Which arrow is closest to the direction of the car’s acceleration while passing the point P?

Concept Question: Circular Motion

An object moves counter-clockwise along the circular path shown below. As it moves along the path its acceleration vector continuously points toward point S. The object

1. speeds up at P, Q, and R.

2. slows down at P, Q, and R.

3. speeds up at P and slows down at R.

4. slows down at P and speeds up at R.

5. speeds up at Q.

6. slows down at Q.

7. No object can execute such a motion.

Next Reading Assignment: W04D2

Young and Freedman: 3.4; 5.4-5.5

Experiment 2: Circular Motion

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