Lecture 13Lecture 13

Circular MotionTorque

Moment of Inertia

Circular Motion

Circular motion is the motion in a circle with constant radius.

Relation to Cartesian coordinates: x = r cos θy = r sin θ

Polar coordinates Polar coordinates (r, θ) are more convenient than Cartesian coordinates to describe circular motion: r = R, only θ = θ(t)

Arch: s = RθDefinition: 1 radian = angle so that s=R 1 revolution = 2π radians θ

y

x s

R

Velocity

Cartesian coordinates:

Polar coordinates: vr =Δ r

Δ t; ω =

Δ θ

Δ tRadial

velocityAngular velocity

(where is in radians/unit time)

For circular motion:

s = R θ ⇒ |Δ s

Δ t| = R|Δ θ

Δ t| ⇒ v = R ω

vr = 0

vx =Δ x

Δ t; vy =

Δ y

Δ t

Angular Velocity and Angular Acceleration

ω = limΔ t→0

ω = limΔ t→0

Δ θ

Δ t

α =Δ ω

Δ t

α = limΔ t→0

α = limΔ t→0

Δ ω

Δ t

EXAMPLE: Two balls

Two balls connected by thin rod as shown, at distances R and 2R from the center, move in circles.

EXAMPLE: Two balls

Two balls connected by thin rod as shown, at distances R and 2R from the center, move in circles.

Same angular speed ω for both (same angle in any Δt)

Different (linear) speeds (ball 2 travels twice the distance in any Δt) :

=1v R=2 2v R

 Uniform circular motion (UCM)

is constant.Angle covered in time interval ∆t: ∆ = ∆t.

If we choose 0 = 0 at t = 0, it’s = t

tx

y

t

x = R cos ty = R sin t

yx

The Cartesian coordinates are sine-functions: 

For any periodic motion

• Period T : Time it takes to go back to the same situation (same position, same velocity).

• Frequency f : Number of revolutions per unit time. Units: Hz (turns per second), rpm (rev per min)

t

y T

2T p

=1f

T= p= 2 f

Example: Ferris Wheel

A Ferris Wheel of radius 8.0 m rotates at a constant rate of 1.5 rpm. Find:

a. The period

b. The linear speed of a cabin.

Example: Ferris Wheel

A Ferris Wheel of radius 8.0 m rotates at a constant rate of 1.5 rpm. Find:

a. The period

b. The linear speed of a cabin.

= = =1 1 0.67 minrev1.5

min

Tf

=v R

ω = 1.5rev

min⋅2 π rad

1 rev⋅1 min

60 s= 0.16

rad

s

= (0.16rad

s ) (8 m ) = 1.3 m/s

ACT: Ferris wheel

The ferris wheel in the figure rotates counterclockwise at a uniform rate. What is the direction of the average acceleration of a gondola as it goes from the top to the bottom of its trajectory?

A. Down

B.

C. The acceleration is 0 because the motion is uniform.

Radial or centripetal acceleration

During uniform circular motion, speed is constant, but velocity is not!!! The direction keeps changing!

(constant

speed) ≠ 0 Perpendicular/normal/radial/centripetal acceleration

Points to the center of the circle

a⃗ = a⃗∥+ a⃗

⊥

Radial acceleration

vf

vi

vf -vi

vf

vivf -vi

Visually: Let’s look at the average acceleration.

a points toward the center

(Ok, it’s sloppy: we should be taking the limit as Δt 0, but you get the idea…→

vf -vi points toward the center

Magnitude of Radial Acceleration

Rv =||Rr =||For UCM:

In UCM, all the acceleration is centripetal. Thus,

RvRa

22

r == Radial acceleration

The subtle point: To have uniform circular motion, the acceleration needs to be exactly v²/R.

a too small; direction changes too little

avi

vf

a = v²/R

avi

vf

a

vi

vf

a too large; direction changes too much

5-8 Satellites and “Weightlessness”The satellite is kept in orbit by its speed – it is continually falling, but the Earth curves from underneath it.

v² =a R

Gravity decides acceleration at certain height.

The right speed is needed to stay at that orbit

Acceleration simulator

In the movie “The Right Stuff”, a seat at the end of a long arm that rotates very fast is used to prepare astronauts for high accelerations.

ω

R If R = 5 m, what is the speed needed to have a = 5g ?

25g R=

5gR

= ~ 0.5 turns/s= √5×9.8 m/s2

5 m= 3.2

1

s×

1 turn

2 π

ACT: Angular speed

A. π rad/s

B. 2π rad/s

C. 4π rad/s

A horizontal rod of length 1 m turns with constant speed about a vertical axis. The blue tip of the rod makes one turn each second. Find the angular speed of the red spot painted in the middle of the rod.

blue tip

red spot

(1 m)(2 rad/s) 6.28 m/s(0.5 m)(2 rad/s) 3.14 m/s

v Rv R

p

p

= = =

= = =

The linear speed is different for the two points:

1 m

It makes sense: The blue tip is covering twice the distance covered by the red spot in the same time.

6.28 m3.14 m

Non-uniform circular motion

ar at

a

R

v

Slowing down

R

v

Speeding up

ar

ata

Radial acceleration (towards the center) changes in direction →

Tangential acceleration (tangential to trajectory) changes in speed→

Non-uniform circular motion

vf - vi does not point to the center, so a does not point to the center either!

vf

vivf - vi

vf

viR

vf - vi

(Sloppy) visual proof, for the slowing down case:

Equations of motion when α=constant

The derivation is formally identical to what we did in 1D:

Example: Grinding wheelAt t = 0, a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0 rad/s2 until a circuit breaker trips at t = 2.00 s. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. What was its acceleration as it slowed down?

Example: Grinding wheelAt t = 0, a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0 rad/s2 until a circuit breaker trips at t = 2.00 s. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. What was its acceleration as it slowed down?

Part 1: Find the angular speed at t = 2.00sω = ω0 + α t = 24.0 rad/s+ (30.0 rad/s2) (2.00 s)

= 84.0 rad/s

Part 2: Find the angular acceleration:ω2 − ω0

2 = 2 α Δθ = 24.0 rad/s+ (30.0 rad/s2) (2.00 s )

⇒ α =ω2 − ω0

2

2 Δθ=

0 − (84 rad/s)2

2 (432 rad )= −8.17 rad/s2

2tanF r mra=

Torque: An intuitive approach

A particle of mass m is constrained by a massless rod of length r to move in circles about point P. A force F is applied on the particle. This is the only force applied on it. What is the angular acceleration of the system?

rF

mFr

Ftan

Ftan produces a tangential acceleration:

tan tanF ma=

I

ta ra=tanF m ra=

τ = Iα

P

Torqueτ = Ftan r

=Fr sinθ

Newton’s second law for rotations

The newton’s 2nd law for rotational motion.

The angular acceleration will be the result of adding all these torques:

τnet=∑ τi=Iα

For each force on a system, the torque depends on the point on which it is applied and its orientation

Moment of Inertia

The moment of inertia is to rotation what mass is to translation.

Indicates how hard it is to rotate an object.

It depends on the position and orientation of the axis.

Units: kg m²

Large I

ω

Small I

ω

DEMO: Bars

I=∑miri2

Example: Moment of inertia of a square of side L made with four identical particles of mass m and four massless rods.

a. For rotations about an axis perpendicular to the square, through the center of the square.

m m

mmAxis

L

Perpendicular distance between each mass and the axis:

rr=√2

2L

I=4 {m(√22

L)2

}

I=2mL2

Example: Moment of inertia of a square of side L made with four identical particles of mass m and four massless rods.

b. For rotations about an axis in the plane of the square, through the center of the square.

m m

mm

Axis

L

Perpendicular distance between each mass and the axis: r

r=L2

I=4 {m(L2

)2

}

I=mL2

Example: Moment of inertia of a square of side L made with four identical particles of mass m and four massless rods.

b. For rotations about an axis in the plane of the square, through one side of the square.

m m

mm

Axis

L

for two of the particles;0 for the other two.

r Lr=

=

Perpendicular distance between each mass and the axis: r

I=2{mL2}

I=2mL2

Recap:Example: Moment of inertia of a square of side L made with four identical particles of mass m and four massless rods.

m m

mm

AxisL

m

m

m

m

Axis

L

m m

mmAxis

L

The moment of inertia depends on the position and orientation of the axis.

I=mL2I=2mL2 I=2mL2

ACT: Triangle

Three identical balls are connected with three identical, rigid, massless rods. The moments of inertia about axes 1, 2 and 3 are I1, I2 and I3. Which of the following is true?

A. I1 > I2 > I3

B. I1 > I3 > I2

C. I2 > I1 > I3 3

2

1

moments of inertia of extended objects

Solid sphere of mass M and radius R, about the axis through its center:

225

I MR=

Solid disk (or cylinder) of mass M and radius R, about the perpendicular axis through its center:

212

I MR=

More: See book or formula sheet for other shapes.

Different Axis —> different IDifferent Shape —> different I