Translational vs Rotational

2

//

1/ 2

Δ====

==

mx

v dx dta dv dtF map mvKE mvWork Fd

2

//

1/ 2

θω θα ωτ α

ω

ωτθ

Δ====

==

I

d dtd dtI

L IKE IWork

2 /

θω

ατ ⊥

Δ = Δ=

====

c

t

s rv ra v ra r

FrL pr

Connection

Table 10.1, p.297

Translational and Rotational Kinematics

In Sum….

Uniform Circular MotionMotion in a circle at constant angular speed.

: angular velocity (rad/s)ω

Rotation Angle

Δθ = rotation angle

Δs = arc length

r = radius

sr

θ ΔΔ =

The rotation angle is the ratio of arc length to radius of curvature.For a given angle, the greater the radius, the greater the arc length.

r

s = ? s = C = 2πr

θ = s/rθ = 2 πr /r

θ = 2 π

How many degrees in 1 radian?

How many radians in 3600 ?Ans: 3600 = 2 π radians

Consider a circle with radius r.

2 π rad = 3600

1 rad = 3600 /2 π = 57.30

d

sθ

r = 4 m

A box is fastened to a string that is wrapped around a pulley. The pulley turns through an angle of 430. What is the distance, d, that the box moves?

43 43 1x=o o

243360

x π= o

o

.75rad=

First: How many radians is 430 ?

d

sθ

r = 4 m

s = r θd = r θd = 4m (0.75 rad)d = 3 m radd = 3 m

Radian is dimensionless and is dropped!

A box is fastened to a string that is wrapped around a pulley. The pulley turns through an angle of .75 rad. What is the distance, d, that the box moves?

430 = .75rad

Angular Velocity: angular velocity (rad/s)ω

θω =ddt

( / )=

d s rdt

1=

dsr dt

tΔ

vr

= ω=v r

: angular velocity (rad/s)v: tangential velocity (m/s)ω

θω = =d vdt r

v rω=

Angular & Tangential Velocity

is the same for every point & v varies with r. ω

ProblemCalculate the angular velocity of a 0.500 m radius car tire when the car travels at a constant speed of 25.0 m/s.

ω =vr

25 /.5m sm

=

50 /rad s=

Insert rad for angular speed. Keep it to define units.

: angular velocity (rad/s)v: tangential velocity (m/s)ω

θω = =d vdt r

v rω=

Angular & Tangential Velocity

is the same for every point & v varies with r. ω

Tangential and Angular Acceleration

ωα =ddt

=tdvadt

( )ω=

d rdt

α=ta r

=drdtω

Centripetal and Tangential Accelerations

2

=cvar

α=ta r2( )ω

=rr

2ω=ca r

(r constant)

Total Acceleration

2 2r ta a a= +

r θa = -a r + a θ̂ˆ

/θ α= = =ta a r dv dt2 2 /ω= = =r ca a r v r

Angular –Tangential BikeA bike wheel with a radius of 0.25 m undergoes a constant angular acceleration of 2.50 rad/s2. The initial angular speed of the wheel is 5.00 rad/s. After 4.00 s

a) What angle has the wheel turned through?b) What is the final angular speed?c) What is the final tangential speed of the bike?d) How far did the bike travel?

20

12

t tθ ω αΔ = + 2 215 / 4 2.5 / (4 ) 402

rad s s rad s s rad= + =

0f tω ω α= + 25 / 2.5 / 4 15 /rad s rad s s rad s= + =

15 / .25 3.75 /v r rad s m m sω= = =

40 .25 10d s r rad m mθ= Δ = Δ = =

Givens:

0

2

5 /

2.5 /4

rad s

rad st s

ω

α

=

==

NEW: Rotational InertiaThe resistance of an object to rotate.

The further away the mass is from the axis of rotation, the greater the rotational inertia.

Rotational InertiaDepends on the axis.

Which has greater Rotational Inertia? (Both have Same m, R)

2=CMI mr

212

=CMI mr

Mass further from axis of rotation.WHY?

For a system of point particles:

Where r is the distance to the axis of rotation.

2= ∑ = ∑i i iI m r I

Calculate Rotational Inertia

SI units are kg.m2

Calculate: Moment of Inertia of a Rigid Object

22= =∫ ∫I r dm R dm

The trick is to write dm in terms of the density:

ρ=dm dV

( )2 2 4

0

12 2

πρ πρ= = =∫ ∫R

I r dm r Lr dr LR

2 π=dV rL drDivide the cylinder into concentric shells with radius r, thickness dr and length L:

212

=zI MR2But /M R Lρ π=

Moments of Inertia of Various Rigid Objects

Superposition of Inertia The Parallel Axis Thm

The theorem states I = ICM + MD 2

I is about any axis parallel to the axis through the center of mass of the object

ICM is about the axis through the center of mass

D is the distance from the center of mass axis to the arbitrary axis

= ∑ iI ISuperposition: Inertia ADD

Moment of Inertia for a Rod Rotating Around One End

The moment of inertia of the rod about its center is

The position of the CM isD=½ L

Therefore,

2112CMI ML=

2CM

22 21 1

12 2 3

I I MD

LI ML M ML

= +

⎛ ⎞= + =⎜ ⎟⎝ ⎠

It is easier to rotate a rod about its center than about an end.

Rotational InertiaWhich reaches the bottom first?

(Same mass and radius)

Why Solid Cylinder?

2=CMI mr

212

=CMI mr

Less Rotational Inertia!

Rolling

The red curve shows the path moved by a point on the rim of the object. This path is called a cycloid

The green line shows the path of the center of mass of the object which moves in linear motion.

Fig. 10.28, p.317

All points on a rolling object move in a direction perpendicular to an axis through the instantaneous point of contact P. In other words, all points rotate about P. The center of mass of the object moves with a velocity vCM, and the point P ’ moves with a velocity 2vCM. (Why 2?)

Rolling without Slipping

For pure rolling motion, (no slipping) as the cylinder rotates through an angle θ, its center moves a linear distance s = Rθ with speed vCM. At any instant, the point on the rim located at point P is at rest relative to the surface since no slipping occurs.

Fig. 10.29, p.318

The motion of a rolling object can be modeled as a combination of pure translation and pure rotation.

Translation: CM moves with vCM.

Rotation: All points rotate about P with angular speed .ω

212

= PK I ω

2 21 12 2

= +CM CMK I mvω

Rolling Without SlippingKinetic Energy

m 2CMUse Parallel-axis Th : = +PI I MR

2 2CM

1 ( )2

= +K I MR ω

The total kinetic energy of a rolling object is the sum of the rotational kinetic energy about the center of mass and the translational kinetic

energy of the center of mass.

Rolling Without SlippingFriction

If an object rolls without slipping then the frictional force that causes the rotation is a static force.If no slipping occurs, the point of contact is momentarily at rest and thus friction is static and does no work on the object. No energy is dissipated and Mechanical Energy is Conserved. If the object slips, then friction is not static, does work on the object and dissipates energy.

2 21 12 2

ω= +CM CMMgh Mv I

=i fE E

2 21 1 ( )2 2

= + CMCM CM

vMv IR

Rolling Without SlippingConservation of Energy

2 2 21 1 2( )( )2 2 5

= + CMCM

vMv MRR

2 21 12 5

= +CM CMMv Mv 2710

= CMMv

107

=CMv gh

212

=Mgh Mv

=i fE E

Rolling Without SlippingConservation of Energy

107

=CMv gh

Compare to Slipping Only:

2=CMv gh

Some of the gravitational potential energy goes into rolling the sphere so the translational velocity of the cm is less when rolling!

What about little ball big ball?

212

=Mgh Mv

=i fE E

Rolling Without SlippingConservation of Energy

107

=CMv gh

Compare to Slipping Only:

2=CMv gh

What about the acceleration down the incline?

Rolling down:

Does ω change while in flight?

10 Pre Lab

Total Kinetic EnergyA 1.0-kg wheel in the form of a solid disk rolls without slipping along a horizontal surface with a speed of 6.0 m/s. What is the total kinetic energy of the wheel?

2 21 12 2

mv Iω= += +total trans rotK K K

2 2 21 1 1( )( )2 2 2

vmv mrr

= +

2 21 12 4

mv mv= +

23 (1 )(6 / )4

kg m s=234

mv= 27J=

What I to use?

Why Solid Cylinder?

2=CMI mr

212

=CMI mr

PROVE IT!

Total Energy

2 20

1 12 2

mgh mv Iω= +

trans rotE KE KE PE= + +

02

2/

mghvm I r

=+

2 21 1 ( )2 2

vmv Ir

= +

2

2/

mghvm I r

=+

0v gh=

043

v gh=

General equation for the total final velocity at the end of the ramp:

Solid Disk has the greatest velocity at the bottom of the ramp! Note: the velocity is independent of the radius!

Torque: Causes Rotations

The moment arm, d, is the perpendiculardistance from the axis of rotation to a line drawn along the direction of the force

sinFr Fdτ φ= =

lever arm: sind r φ=

The horizontal component of F (F cos φ) has no tendency to produce a rotation

Torque: Causes RotationssinFr Fdτ φ= =

The direction convention is:

Counterclockwise rotations are positive.

Clockwise rotations are negative.

TorqueIs there a difference in torque?

(Ignore the mass of the rope)

NO! In either case, the lever arm is the same!

What is it? 3m

If the sum of the torques is zero, the system is in rotational equilibrium.

Newton’s 1st Law for Rotation

250 3 750girl N m Nmτ = ⋅ = +

500 1.5 750boy N m Nmτ = ⋅ = −

0τ∑ =

Newton’s 2nd Law for Rotation

Iτ α∑ =

The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration.

The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.

Inertia thing

Acceleration thing

Force thing

Net Torque 24.3.314

m kgR m==) ?

) ?ab

τ

α

=

=∑

1 2

( 90 125 ).31411

F r F rN N m

Nm

τ = +

= + −= −

∑

2= 1 mR

2Iτ τα ∑ ∑=

2

11=1/2 24.3kg(.314m)

Nm−⋅

2= 9.2 /rad s−

Net Torque: You Try It!

( 20 )(.5 ) (35 cos30)(1.10 )= − +N m N m

1 1 2 2τ = +∑ F d F d

( 20 )(.5 ) (35 )(1.10 sin 60)= − +N m N mF and d must be mutually perpendicular!

sin

OR

23.3 CCW= + Nm

Fr Fdτ φ= =

IIττ α α= → =

20

1 2 22 /

t t tI

θ θθ ω αα τΔ Δ

Δ = + → = =

20

1:2

Use t tθ ω αΔ = +

A 50 N m torque acts on a wheel of moment of inertia 150 kg m2.If the wheel starts from rest, how long will it take the wheel to make a quarter turn (90 degrees)?

2

2 / 2 3.150 /150

radt sN m kg m

π⋅= =

⋅ ⋅

A tennis ball starts from rest and rolls without slipping down the hill. Treat the ball as a solid sphere. Find the range x.

Rolling Ball Problem

2 2 21 1 2( )( )2 2 5

= +vmv Mrr

2 21 12 2

ω= +CM CMmgh mv I

107

=v gh

cosxR v t v tθ= =

210 sin2

v t gtθ= −2 sinvt

gθ

∴ =

cosxR v t v tθ= =2 2sin cosv

gθ θ

= (10 / 7 )2sin cosghg

θ θ=

10 sin 27

hRg

θ= = .25m

212y yy v t a tΔ = +

Does ω change while in flight?

A meter stick is attached at one end (the zero mark) and is free to rotate on a horizontal, frictionless table. A particle of mass 0.400 kg is shot at the meter stick with initial speed 3.00 m/s, as shown. The particle strikes and sticks to the meter stick at the 75.0-cm mark. The meter stick has a mass 0.100 kg.a) Calculate the rotational inertia and the center of mass of the stick/particle system.

b) Calculate the angular speed of the system just after the particle hits and sticks to the meter stick.

c) How long does it take for the system to make one complete revolution?