4.1 Rotational kinematics4.2 Moment of inertia4.3 Parallel axis theorem 4.4 Angular momentum and

rotational energy

CHAPTER 4: ROTATIONAL MOTION

Part 1

Rotational kinematics

Riview of rotations• Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds.• Klyde’s angular velocity is:

(a)(a) the same as Bonnie’s (b)(b) twice Bonnie’s(c)(c) half Bonnie’s

REVIEW of ANGLE VELOCITY• The angular velocity of any point on a solid object rotating about a fixed axis is the same.

– Both Bonnie & Klyde go around once (2pi radians) every two seconds.

Their “linear” speed v will be different since v = r.

BonnieKlyde V21V

How about their “linear” speed v ?The same or different? ?

Review: Rotational Variables.

• Rotation about a fixed axis:– Consider a disk rotating about

an axis through its center:

• First, recall what we learned aboutUniform Circular Motion:

(Analogous to )

dtd

dtdxv

Rotational Variables...

• Now suppose can change as a function of time:• We define the angular acceleration:

2

2

dtd

dtd

Consider the case when

is constant. We can integrate this to find and as a function of time:

t0 constant

200 t

21t

Rotational Variables...

• Recall also that for a point at a distance R away from the axis of rotation:– x = R (distance in the circle)– v = R

And taking the derivative of this we find:– a = R

R

v

x

t0 constant

200 t

21t

Example: Wheel And Rope

• A wheel with radius R = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4 m/s2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2 radians)

aa

R

Solution

• Use a=R to find : = a / R = (4 m/s2 )/ 0.4 m = 10 rad/s2

• Now use this equations just as you would use the kinematic equations from the beginning of the semester.

times80radrot

21 xrad 500rev

200 t

21t = 0 + 0(10) + (10)(10)2 = 500 rad

21

aa

R

Part 2

Moment of Inertia

Rotation & Kinetic Energy

• Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods).

• The kinetic energy of this system will be the sum of the kinetic energy of each piece:

rr1

rr2rr3

rr4

m4

m1

m2

m3

Compute: Kinetic Energy of Rotation system

• So: but vi = ri K m vi ii

12

2

K m r m ri ii

i ii

12

12

2 2 2

which we write as:

K 12

2I

I m ri ii

2

Define the moment of inertiamoment of inertiaabout the rotation axis

rr1

rr2rr3

rr4

m4

m1

m2

m3

vv4

vv1

vv3

vv2

I has units of kg m2.

Rotation & Kinetic Energy...

• The kinetic energy of a rotating system looks similar to that of a point particle:

Point ParticlePoint Particle Rotating SystemRotating System

I m ri ii

2

K 12

2I K mv12

2

v is “linear” velocitym is the mass.

is angular velocityI is the moment of inertiaabout the rotation axis.

Moment of Inertia

• Notice that the moment of inertia I depends on the distribution of mass in the system.• The further the mass is from the rotation axis, the bigger the moment of inertia.

* For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass).* In rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics!

K 12

2I I m ri ii

2 So where

Calculating Moment of Inertia

• We have shown that for N discrete point masses distributed about a fixed axis, the moment of inertia is:

I m ri ii

N2

1where ri is the distance from the mass i to the axis of rotation.

Example: Calculate the moment of inertia of four point masses (m) on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square:

mm

mm

L

Calculating Moment of Inertia...

• The squared distance from each point mass to the axis is:

2L

2L2r

222

mm

mm

Lr

L/2

2Lm4

2Lm

2Lm

2Lm

2LmrmI

22222N

1i

2ii

so

I = 2mL2

Using the Pythagorean Theorem

Learning check

• Now calculate I for the same object about an axis through the center, parallel to the plane (as shown):

mm

mm

L

r

4Lm4

4Lm

4Lm

4Lm

4LmrmI

22222N

1i

2ii

I = mL2

Calculating Moment of Inertia...

• Finally, calculate I for the same object about an axis along one side (as shown):

mm

mm

L

r

2222N

1i

2ii 0m0mmLmLrmI

I = 2mL2

Calculating Moment of Inertia...

• For a single object, I clearly depends on the rotation axis!!

L

I = 2mL2I = mL2

mm

mm

I = 2mL2

Check: Moment of Inertia

• A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is Ia, Ib, and Ic respectively.– Which of the following is correct:

(a)(a) Ia > Ib > Ic

(b)(b) Ia > Ic > Ib

(c)(c) Ib > Ia > Ic

a

b

c

Solution

a

b

c

m

m m

L

L

Masses m and lengths L as show:

Ia m L m L mL 2 2 82 2 2

Calculate moments of inerta:

Ib mL mL mL mL 2 2 2 23

Ic m L mL 2 42 2

So (b) is correct: Ia > Ic > Ib

Calculating Moment of Inertia... For a continuous solid object

• For a discrete collection of point masses we found:

• For a continuous solid object we have to add up the mr2 contribution for every infinitesimal mass element dm.

• We have to do anintegral to find I :

I m ri ii

N2

1

r

dm

I r dm2

Learn by heartMoments of Inertia

• Some examples of I for solid objects:Thin hoop (or cylinder) of mass M and radius R, about an axis through its center, perpendicular to the plane of the hoop.

I MR 2

R

Thin hoop of mass M and radius R, about an axis through a diameter.

I 12

2MRR

2222 MRdmRdmRdmrI

Moments of Inertia...

• Some examples of I for solid objects:

Solid sphere of mass M and radius R, about an axis through its center.

I 25

2MR

R

I 12

2MR

RSolid disk or cylinder of mass M and radius R, about a perpendicular axis through its center.

Moment of Inertia

• Two spheres have the same radius and equal masses. One is made of solid aluminum, and the other is made from a hollow shell of gold.– Which one has the biggest moment of inertia about an axis through its center?

same mass & radius

solid hollow

(a) solid aluminum (b) hollow gold (c) same

Hint

• Moment of inertia depends on mass (same for both) and distance from axis squared, which is bigger for the shell since its mass is located farther from the center.– The spherical shell (gold) will have a bigger moment of inertia.

same mass & radius

ISOLID < ISHELL

solid hollow

Moments of Inertia...

• Some examples of I for solid objects (see also Tipler, Table 9-1):

Thin rod of mass M and length L, about a perpendicular axis through its center.

I 1

122ML

L

Thin rod of mass M and length L, about a perpendicular axis through its end.

I 13

2ML

L

Part 3

Parallel Axis Theorem

Parallel Axis Theorem

• Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass, ICM, is known.• The moment of inertia about an axis parallel to this axis but a distance D away is given by:

IPARALLEL = ICM + MD2

So if we know ICM , it is easy to calculate the moment of inertia about a parallel axis.

Parallel Axis Theorem: Example

• Consider a thin uniform rod of mass M and length D. Figure out the moment of inertia about an axis through the end of the rod.

IPARALLEL = ICM + MD2

ICM ML1

122We know

IEND ML M L ML

112 2

13

22

2So

which agrees with the result on a previous slide.

L

D=L/2 Mx

CM

ICMIEND

Part 4

Angular momentum and rotational energy

Angular momentum

Defination:

CM2

ii

i

ii

iii

i

Irm

])vm[xr(LLPxrL

Example: compute L for the system m=100g, a=10cm, =(2/5) rad/s

a

mL=2ma2. =2x0.1x 10-4 x0.1

=2 10-6.

r

v

L

Complete Motion by linear + rotation

• The total kinetic energy of a system of particles include 2 parts

2CM

2iiNET MV

21um

21K

KR KCM For a solid object rotating about its center of mass, we

now see that the first term becomes:

2CM

2CMTOT MV

21I

21K

2iiR um

21K Substituting ii ru

2ii

2R rm

21K but CM

2ii rm I

Connection with CM motion...

• So for a solid object which rotates about its center or mass and whose CM is moving:

2CM

2CMNET MV

21I

21K

VCM

We will use this formula more in coming lectures.

Similarity Between Linear and Rotational MotionsAll physical quantities in linear and rotational motions show striking similarity.

Quantities Linear RotationalMass Mass Moment of Inertia

Length of motion Distance Angle (Radian)Speed

AccelerationForce Force TorqueWork Work WorkPower

MomentumKinetic Energy Kinetic Rotational

2I mr

rvt

t

vat

t

maF IcosW Fd

vFP P

2

21 mvK 2

21 IKR

LM

W

vmp IL

Review of today’s lecture

• Rotational Kinematics – Analogy with one-dimensional kinematics

• Kinetic energy of a rotating system – Moment of inertia – Discrete particles Continuous solid objects

• Parallel axis theorem

Problem

Use the law of energy conservation to compute the velocity of the ball when it hits to the ground.

Hint : There are 2 motions

(frictionless case)

H=3m

V0=0, 0=0

M=2kg, r=4cm

A=45 Deg

Change the ball by Cylinder have the same M and r