rotational motion & angular momentum · 2018-03-02 · p.o.d. 5: the blades of an electric...

Rotational Motion &

Angular Momentum

Rotational Motion

Every quantity that we have studied with

translational motion has a rotational

counterpart

TRANSLATIONAL ROTATIONAL

Displacement x Angular Position

Velocity v Angular velocity

acceleration a Angular acceleration

Mass m Inertia I

Momentum p Angular Momentum L

Force F Torque

Angular Position

r

s

Arclength

Radius (from axis of

rotation)

Measured in radians – all angular quantities will be measured in

radians (NOT degrees)

• In translational motion,

position is represented by a

point, such as x.

• In rotational motion,

position is represented

by an angle, such as ,

and a radius, r.

x

linear

0 5

x = 3

r

0

p/2

p

3p/2angular

Angular Position

Angular

Displacement

• Linear displacement

is represented by the

vector Dx.

• Angular displacement is represented by D, which is not a vector, but behaves like one for small values.

x

linear

0 5

Dx = 4

D 60

0

p/2

p

3p/2angular

Angular Displacement

0DWhich direction is positive (by

convention)?

Positive – Counterclockwise

Negative - Clockwise

Compare to

D𝒙 = 𝒙 − 𝒙𝒐

Tangential vs. angular displacement

• A particle that rotates through an angle Dalso translates through a distance s, which is the arc length defining its path.

• This distance s is related to the angular displacement D by the equation s = rD, as we saw in a previous slide.

rD

ss

D

EXAMPLE: (a) What is the angular position, , if we go

around a circle two times?

Ans. = s

r=

2(2p𝑟)r

= 4π

(b) Say you go a quarter turn more, what is D?

Ans. D = o = 9𝜋

2− 4π =

𝜋

2(c) What is the arclength covered in total between

Probls. (a) and (b) if the radius of the circle is 3 m?

Ans. s = r = (3 m)(4p + 𝜋

2) =3m(

9𝜋

2) =

27𝜋

2m

P.O.D. 1: Two

synchronous

communications

satellites are put into an

orbit whose radius is

r = 4.23 x 107 m. The

two adjacent satellites

have an angular

separation of 2. Find

the arc length, s, that

separates the satellites.

Speed and velocity

• The instantaneous velocity of a particle in a circular path has

magnitude vT = 𝐝𝐬

𝐝𝐭and is tangent to

the circle.

• The same particle rotates with an avg. angular velocity

= D𝛉

D𝐭=

𝛉𝒇−𝛉𝒊

D𝐭• or instantaneous angular velocity

= 𝐝𝛉

𝐝𝐭• Tangential and angular speeds are

related by the equation

vT = r .

rD

s

vT

vT

rr

Angular Velocity

tavg

D

D

dt

dinst

Units – rad/s, rev/s

Direction – Same as displacement

(positive is counterclockwise, negative

is clockwise)

Magnitude – Angular Speed

Compare to

𝒗𝒂𝒗𝒈 =∆𝒙

∆𝒕and

𝒗𝒊𝒏𝒔𝒕 = 𝒅𝒙

𝒅𝒕

Derivation of tangential velocity formula

rv

rdt

d

dt

ds

rs

Take the derivative of each

side of s = r with respect

to t.

Substituting = 𝐝𝛉

𝐝𝐭and vT =

ds

dt

Derivation of tangential velocity formula

rvT

•OBSERVATIONS:

•All points on object have the same angular speed,

•Those points farther from the center have a greater

linear (tangential) speed, vT

•These points have to travel farther in the same time

tavg

D

D

P.O.D. 2: A gymnast on a high bar swings

through two revolutions in a time of 1.90 s.

(a) Find the average

angular velocity (in

rads/s) of the gymnast.

(b) What is her angular

displacement (in rad)

after t = 1 s.

(c) Find her tangential

velocity if the gymnast

is 1.8 m tall.

Acceleration• Tangential acceleration is given by

aT = 𝐝vT𝒅𝒕

• Instantaneous angular

acceleration of this particle

is given by

= 𝐝𝒅𝒕

• Average Angular acceleration of this

particle is given by

= DD𝒕

• Tangential and angular accelerations

are related by the equation

aT = r

rD

s

aT

aT

r

Angular Acceleration

• Angular Acceleration is how

the angular velocity changes

with time

• This is not the centripetal

acceleration – that tells about a

translational acceleration

tavg

D

D

dt

dinst

Units – rad/s2

Compare to

𝒂𝒂𝒗𝒈 =∆𝒗

∆𝒕and

𝒂𝒊𝒏𝒔𝒕 = 𝒅𝒗

𝒅𝒕

Sample Problem

A DVD spins at an angular velocity of 3 rad/s and 5

seconds later is spinning at a velocity of 8 rad/s. Its

radius is 0.02 m. Find

(a) Angular acceleration,

=∆𝜔

∆𝑡=𝜔𝑓−𝜔𝑖

∆𝑡=

(a) Tangential acceleration, aT

(a) Centripetal acceleration, ac, at your final angular

velocity

P.O.D. 3: A jet awaiting clearance for takeoff is

momentarily stopped on the runway. The fan blades are

rotating with an angular velocity of 110 rad/s, where the

negative sign indicates a clockwise rotation. As the

plane takes off, the angular velocity of the blades

reaches 330 rad/s in a time of 14 s.

(a) Find the angular acceleration, assuming it to be

constant.

(b) Find the tangential acceleration, if the fan blade has

a radius of 1.5 m.

Centripetal Acceleration

rr

vac

22

A pendulum is swinging back and

forth.

At the bottom of the swing the

force of gravity is pulling it

downwards but it doesn’t fall down.

This means there must be a force

pulling upwards to balance it out.

This is the centripetal force.

Since F = ma, the center-seeking

acceleration is called centripetal

acceleration and is given by:

Fc

Fg

Derivation of

Acceleration

ra

rdt

dv

rdt

d

dt

dv

rv

T

T

T

T

Tangential Acceleration

Take the derivative of each side of

v = r with respect to t.

Substituting = 𝐝𝐝𝐭

Substituting a = dv

dt

Tangential Acceleration vs. Angular Acceleration

vs. Centripetal Acceleration

• The net acceleration is the sum

of the tangential and

centripetal accelerations.

2 2

r ta a a

rr

vaa 2

2

cr

aT = r

Sample Problem

A compact disk rotates about an axis

through its center according to

t6t3

1)t( 3

(a) Determine its angular velocity and angular

acceleration at time t = 5 seconds.

Ans. To find the angular velocity, take the first derivative.

To find the angular acceleration, take the second derivative

srads196)5(6t

dt

d)t(' 22

2srads10)5(2t2

dt

d)t(''

Sample Problem (Cont.)

A compact disk rotates about an axis

through its center according to

t6t3

1)t( 3

(a) What is the linear speed of a point 20 cm from the

center at t = 5 s?

(b) What is the linear acceleration at 0.5 m at t = 5 s?Ans.

To find the linear speed, use the relationship, where (5) we

obtained from the previous slide:

v = r = (19 rad/s)(0.20 m) = 3.8 m/s.

To find the linear acceleration, use the relationship

ac = 2r = (19 rad/s)2(0.20 m) = 72.2 m/s2

P.O.D. 4: The gyroscope of a plane is spinning

according to 𝜽 𝒕 = 𝟑𝒕𝟑 + 𝟔𝒕𝟐 ½t.

(a) Find the average angular velocity of the gyroscope between t

= 2 s and t = 5 s.

(b) Find the instantaneous angular acceleration at t = 2 s.

(c) Find the linear speed, vT, at t =

2 s of a point on the gyroscope

if it has a radius of 0.05 m.

(d) Find the linear (tangential)

acceleration, aT, at t = 2 s of

the gyroscope.

(e) Find the centripetal

acceleration, ac, of a point on

the edge of the gyroscope.

Constant Angular Acceleration

• Our kinematics equations have angular

equivalents

• Just as with their linear counterparts, these only

work for constant acceleration

First Kinematic Equation

• v = vo + at (linear form)

– Substitute angular velocity for linear velocity.

– Substitute angular acceleration for linear

acceleration.

• = o + t (angular form)

Second Kinematic Equation

• x = xo + vot + ½ at2 (linear form)

– Substitute angle for position.

– Substitute angular velocity for linear velocity.

– Substitute angular acceleration for linear

acceleration.

• = o + ot + ½ t2 (angular form)

Third Kinematic Equation

• v2 = vo2 + 2a(x xo)

– Substitute angle for position.

– Substitute angular velocity for linear velocity.

– Substitute angular acceleration for linear

acceleration.

• 2 = o2 + 2( o)

Sample Problem

An automobile starts from rest and for

20.0 s has a constant linear

acceleration of 0.8 m/s2 to the right.

During this period, the tires do not slip.

The radius of the tires is 0.330 m. At

the end of the 20.0 s interval what is

the angle through which each wheel

has rotated?

Ans. The angular acceleration can be found from the formula

a = r 𝜶 =𝒂

𝒓 𝜶 =

𝟎.𝟖𝒎

𝒔𝟐

𝟎.𝟑𝟑𝟎 𝒎= 2.42 rad/s2

The angular acceleration should be negative because the tire

spins clockwise.

To find the angular displacement:

(t) = ot + ½t2 = 0(20 s) + ½(2.42 rad/s2)(0.20 s)2 = 484 rad

P.O.D. 5: The blades of an electric

blender are whirling with an angular

velocity of +375 rad/s while the puree

button is pushed in. When the blend

button is pressed, the blades

accelerate and reach a greater angular

velocity after the blades have rotated

through an angular displacement of

+44.0 rad (seven revolutions). The

angular acceleration has a constant

value of +1740 rad/s2.

(a) Find the final angular velocity

of the blades.

(b) Find the angular displacement

of the blades after 10

seconds.

(c) Find the change in tangential

velocity of the blades from the

puree to the blend position if

they have a radius of 0.02 m.

ANGULAR

MOMENTUM!, INERTIA,

ETC.

Angular Momentum

v

m

Angular momentum depends on linear momentum and the distance

from a particular point. It is a vector quantity with symbol L. If rand v are then the magnitude of angular momentum w/ resp. to

point Q is given by L = rp = mvr. In this case L points out of the

page. If the mass were moving in the opposite direction, L would

point into the page. The SI unit for angular momentum is the

kg m2 / s. (It has no special name.)

Angular momentum is a conserved

quantity. A torque is needed to change L,

just a force is needed to change p.

Anything spinning has angular has angular

momentum. The more it has, the harder it

is to stop it from spinning.

Q

r

Angular Momentum: General Definition

If r and v are not then the angle between

these two vectors must be taken into account. The

general definition of angular momentum is given

by a vector cross product:

L = r pThis formula works regardless of the angle. From cross products, the magnitude of the

angular momentum of m relative to point Q is: L = r p sin = m v r. In this case, by the

right-hand rule, L points out of the page. If the mass were moving in the opposite

direction, L would point into the page.

r

v

m

Q

Moment of Inertia vs. Angular Momentum

Any moving body has inertia. (It wants to keep moving at constant

v.) The more inertia a body has, the harder it is to change its linear

motion. Rotating bodies possess a rotational inertia called the

moment of inertia, I. The more rotational inertia a body has, the

harder it is change its rotation. For a single point-like mass w/ respect

to a given point Q, I = mr 2.

I = mr 2

m

r

For a system, I = the sum of each mass

times its respective distance from the

point of interest.

r2

m2

r1

m1

I = mi ri2 = m1 r1

2 + m2r22

Q

Q

Moment of Inertia of various shapes

Moment of Inertia ExampleTwo merry-go-rounds have the same mass and are spinning with the

same angular velocity. One is solid wood (a disc), and the other is a

hollow metal ring. Which has a bigger moment of inertia relative to

its center of mass?

mm

r

r

Ans. I is independent of the angular speed. Since their

masses and radii are the same, the ring has a greater

moment of inertia. This is because more of its mass is

farther from the axis of rotation. Since I is bigger for the

ring, it would more difficult to increase or decrease its

angular speed.

Torque & Angular Acceleration

Newton’s 2nd Law, as you know, is Fnet = ma

The 2nd Law has a rotational analog: net = I

A force is required for a body to undergo acceleration. A “turning force”

(a torque) is required for a body to undergo angular acceleration.

The bigger a body’s mass, the more force is

required to accelerate it. Similarly, the

bigger a body’s rotational inertia, the more

torque is required to accelerate it angularly.

Both m and I are measures of a body’s

inertia

(resistance to change in motion).

Example: The torque of an Electric Saw Motor

The motor in an electric saw brings the circular

blade up to the rated angular speed of 80.0 rev/s

in 240.0 rev. One type of blade has a moment of

inertia of 1.41 x 10-3 kgm2. What net torque

(assumed constant) must the motor apply to the

blade?

Ans. First we need to convert our values into rad/s for calculation

purposes.

o t240 rev x 2p = 1508 rads ? 80 revs/s x 2p = 503 rad/s 0 rad/s ?

We can find the angular acceleration from 2 = o2 + 2

Solving for : = 2−o

2

2= (503 rad/s)2−02

2(1508 rad/s)= 83.89 rad/s2

= I = 1.41 x 10-3 kgm2 83.89 rad/s2 = 0.118 Nm

P.O.D. 6: A Chinese star of mass 0.025 kg and radius 0.03 m is

thrown by Dwight from The Office at his adversary, Jim Halpert.

The Chinese star is thrown from rest. If its final angular velocity is

15 revs/s after 3 sec,

(a) find the angular acceleration of the Chinese star (in rad/s2).

(b) Find the torque of the Chinese star (Assume the Chinese star is

hoop-shaped).

Linear Momentum vs. Angular Momentum

If a net force acts on an object, it must accelerate, which means its

momentum must change. Similarly, if a net torque acts on a body, it

undergoes angular acceleration, which means its angular momentum

changes. Recall, angular momentum’s magnitude is given by

L = mvr

r

v

m

So, if a net torque is applied, angular velocity must

change, which changes angular momentum.

Proof: net = r Fnet = r m a

= r mDvt

= DLt

So net torque is the rate of change of angular momentum, just as net

force is the rate of change of linear momentum.continued on next slide

(if v and r are perpendicular)

Linear & Angular Momentum (cont.)

Here is yet another pair of similar equations, one linear,

one rotational. From the formula v = r , we get

L = mv r = m ( r) r = m r2 = I

This is very much like p = mv, and this is one reason I is

defined the way it is.

In terms of magnitudes, linear momentum

is inertia times speed, and angular

momentum is rotational inertia times

angular speed.

L = I

p = m v

Comparison: Linear & Angular Momentum

Linear Momentum, p

• Tendency for a mass to continue

moving in a straight line.

• Parallel to v.

• A conserved, vector quantity.

• Magnitude is inertia (mass)

times speed.

• Net force required to change it.

• The greater the mass, the greater

the force needed to change

momentum.

Angular Momentum, L

• Tendency for a mass to continue

rotating.

• Perpendicular to both v and r.

• A conserved, vector quantity.

• Magnitude is rotational inertia

times angular speed.

• Net torque required to change it.

• The greater the moment of

inertia, the greater the torque

needed to change angular

momentum.

Example: Spinning Ice SkaterSuppose Mr. Stickman is sitting on a stool that swivels holding a pair of

dumbbells. His axis of rotation is vertical. With the weights far from that axis,

his moment of inertia is 600 kgm2 and he is spinning at an angular velocity of

20 rad/s. When he pulls his arms in as he’s spinning, the weights are closer to the

axis, so his moment of inertia gets to

400 kgm2. What will be his angular

velocity at this moment?

Ans.

I = L = I 600 kgm2(20 rad/s) = 400 kgm2

12,000 = 400

30 rad/s =

P.O.D. 7: An artificial satellite

(m = 1500 kg) is placed into an

elliptical orbit about the earth.

Telemetry data indicate that its

point of closes approach (called

the perigee) is rp = 8.37 x 106 m

from the center of the earth,

while its point of greatest

distance (called the apogee) is

rA = 25.1 x 106 m from the

center of the earth. The speed

of the satellite at the perigee is

vp = 8450 m/s.

(a) Find its speed vA at the

apogee.

(b)Find its angular momentum

at any point in its orbit.

Rotational Kinetic Energy

• A particle in a rotating object has rotational kinetic energy:

Ki = ½ mivi2, where vi = i r (tangential velocity)

2 2

2 2 2

1

2

1 1

2 2

R i i i

i i

R i i

i

K K m r

K m r I

The whole rotating object has a rotational kinetic energy given by:

Rotational Kinetic

Energy example:

A thin walled hollow cylinder (mass

= mh, radius = rh) and a solid

cylinder ( mass = ms, radius = rs)

start from rest at the top of an

incline. Both cylinders start at the

same vertical height ho. All heights

are measured relative to an

arbitrarily chosen zero level thatpasses through the center of mass of a cylinder when it is at the

bottom of the incline. Ignoring energy losses due to retarding

forces, determine which cylinder has the greatest translational

speed upon reaching the bottom.

Rotational Kinetic Energy example (cont.):

Ans. At the top of the incline the cylinder have only gravitational potential

energy. At the bottom of the incline this energy has converted into

translational kinetic and rotational kinetic energy.

Ein = Eout

GPEin = TKEout + RKEout

mgh = ½mvf2 + ½If

2

The angular velocity can be related to the linear velocity vf by f = 𝐯𝐟

𝐫

Substituting the given values and for the angular velocity:

mhgho = ½mhvf2 + ½I(

𝐯𝐟

𝐫𝒉)2

For the hollow cylinder, the moment of inertia is given by: I = mr2

Substituting: mhgho = ½mhvf2 + ½(mhrh

2)(vf

rℎ)2

Simplifying: gho = ½vf2 + ½rh

2(𝐯𝐟

𝐫𝒉)2

gho = ½vf2 + ½rh

2𝐯𝒇𝟐

𝒓𝒉𝟐

gho = ½vf2 + ½𝐯𝒇

𝟐

gho = vf2

𝐠𝐡𝐨= vf

Rotational Kinetic Energy example (cont.):

Ans. At the top of the incline the cylinder have only gravitational potential

energy. At the bottom of the incline this energy has converted into

translational kinetic and rotational kinetic energy.

Ein = Eout

GPEin = TKEout + RKEout

mgh = ½mvf2 + ½If

2

The angular velocity can be related to the linear velocity vf by f = 𝐯𝐟

𝐫

Substituting the given values and for the angular velocity:

msgho = ½msvf2 + ½I(

𝐯𝐟

𝐫𝒔)2

For the solid cylinder, the moment of inertia is given by: I = ½mr2

Substituting: msgho = ½msvf2 + ½(½msrs

2)(vf

r𝑠)2

Simplifying: gho = ½vf2 + ½ ½rs

2(𝐯𝐟

𝐫𝒔)2

gho = ½vf2 + ¼rs

2𝐯𝒇𝟐

𝒓𝒔𝟐

gho = ½vf2 + ¼𝐯𝒇

𝟐

gho = ¾vf2

𝟒

𝟑𝐠𝐡𝐨= vf

PROBLEM 8: