Angular Momentum BFDBFDBFD FND FN

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Course Handout

1. (i) Angular Momentum:

10.1 Angular Momentum of a particle

10.2 Systems of particles

10.3 Angular Momentum and Angular Velocity

10.4 Conservation of angular momentum

10.5 The spinning top

10.6 Review of rotational dynamics

(1) Physics: Resnick, Halliday, Krane vol 1 (5th edition)

Ref. Book: Fundamental of Physics: Halliday, Resnick, and Walker

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10 -5 The spinning topThe figure (a) shows a top spinning about the axis of symmetry.

What is interesting is that if the top spins rapidly

the axis rotates about the z axis, sweeping out a cone (fig. b).

The motion of the axis about the vertical is known asprecessional motion. This is relatively slow

in comparison to the spin motion of the top.

Question: why the top does not fall? Bottom part of

the

top is assumed to be fixed at the origin O. At thecenter of mass (CM) a force Mg gives a torque about

O: =Mg rsin. As the top is spinning, it has an

angular momentum L, directed along its symmetry

axis.

The motion of this symmetry axis about the z axis

(precessional motion) occurs because the torque

produces a change in the direction of the symmetry axis.

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The spinning top

n is the normal force acting upward at the pivot

point o. This normal force produces no torque

about o, as it passes through it.

= rx Mg = Mg rsinand is to the plane

formed by rand Mg.

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A simplified diagram, with the top replaced

by a particle of mass M located at the

tops centre of mass.

z

r

M

Mg

is to the axis of the top andso to the angular momentum L.


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The spinning top

= rx Mg = Mg rsin and is to the planeformed by rand Mg.

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z

r

M

Mg

is to the axis of the top andso to the angular momentum L.

So can change the direct

ion

of L , but not its magnitudersin

The change in L in a small incrementof time dtis given by dL = dtand in thesame direction of

L L + dL


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Figure b illustrates the resulting precessional

motion of the symmetry axis of the top. In a time

t, the change in angular momentum is L = Li - Lf

= t. Because L is to L, the magnitude ofLdoes not change.

Rather, what is changing is the direction of L.

Because the change in angular momentum L is

in the direction of, which lies in thexyplane, the

top undergoes precessional motion.

z

L LL +L

d

Lsin

In a time dt, the axis rotates through an angled and thus the angular speed of precession:

P = d/dt

d = dL/Lsin = dt /LsinP = /Lsin = Mgrsin/ Lsin = Mgr/L

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The tip of the vectorL and the

axis of the top trace out a circle

about the z axis.


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z

LL

L +L

d

Lsin

In a time dt, the axis rotates through an angle d and

thus the angular speed of precession:

P = d/dt

d = dL/Lsin = dt /LsinP = /Lsin = Mgrsin/ Lsin = Mgr/L

The precessional speed is inversely proportional to

the angular momentum and thus rotational angularspeed. The faster the top spins, slower the precession.

=Px L

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L=I

P = Mgr/L magnitude of ,P and L


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Problem: 10.26

A top is spinning at 28.6 rev./s about an axis making an angle of

34.0 with the vertical. Its mass is 492 g and its rotational inertia is

5.12x 10-4 kg.m2. The center of mass is 3.88 cm from the pivot point.The spin is clockwise as seen from the above. Find the magnitude

(in rev/s) and direction of the angular velocity of precession.

P = Mgr/I

= (0.492 kg)(9.81 m/s2)(3.88 x 10-2m) / {(5.12 x 10-4 kg .m2)

(2 28.6 rad/s)}

= 2.04 rad/s = 2.04 rad/s /(2) = 0.324 rev /s.

Time period (T)= time for one rotation or revolution

2 in T sec. = 2/T = 2*n, n= no. of revolution/sec; T = 1/n sec

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Translational Quantity Rotational Quantity

Velocity v = dr/dt Angular velocity = d/dt

Acceleration a = dv/dt Angular

acceleration

= d/dt

Mass m Rotational inertia I=mr2

Force Fext Torque = r x F

Newtons second

law

Fext = ma Newtons second

law for rotation

about a fixed axis

ext= I z

Momentum of a

particle

p= mv Angular

momentum of a

particle

l= r x p

Review of translational and rotational dynamics

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Translational Quantity Rotational Quantity

Momentum of a

system of particle

P = MVcm Angular

Momentum of a

system of particle

L = I

General form of

Newtons secondlaw

Fext= dP/dt General form of

Newtons secondlaw of rotation

ext= dL/dt

Conservation of

momentum in a

system of

particles for

which Fext= 0

P= pn =constant

Conservation of

angular

momentum in a

system of

particles for

which ext= 0

L= ln = constant

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Problem:

A uniform stick of mass M and length L is suspended

horizontally. The end B of the stick is attached to the edge of atable and the end A is held by hand. At the instant when end A

of the stick is suddenly released, find

(a)Torque about B.

(b)Angular acceleration about B.

B A = MgL/2

= I MgL/2 = (ML2/3) =3g/2L

Mg

cm

= r x F

= r F sin = r ma

= r m ( r) = mr2 = I

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r= (6 i + 5.0 tj) m; v =d (r)/dt = 5.0j m/s.

L = rx p = (6 i + 5.0 tj) m x (2 kg) x 5.0j m/s = 60 k kg.m2/s

Class work:

The position vector of a particle of mass 2.0 kg is given as a function of time

by r= (6 i + 5.0 tj) m. Determine the angular momentum of the particle aboutthe origin as a function of time.

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Rotation is caused by torque which

is defined as Fr

The unit of torque is Nm.

m

F

r

rF

tF

o

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R.f

f

R.f f

R.f

= 2Rf

F = 0

= 0F = 2f

= Rf

F = f

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A disk of mass M and radius R rotates about a horizontal axis

through its center with angular speed o. (a) What is the

angular momentum? (b) A chip of mass m breaks off the edge

of the disk at an instant such that the chip rises vertically

above the point at which it broke off. How high above the point

does it rise before starting to fall?

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(b) The initial speed of the chip = vo = 0 R.

v=u gt = vo

gt =0

The chip decelerates in a time t = vo/g, and during this time the

chip travels a distance of h v2 =u2 2g hmax hmax =u2

/2 g = vo2 /2 g = 0

2R2/(2g)

(a) L = I0 = (1/2)MR20 .

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Class test on next Monday 26th August ,2013

Time: 11:30 12:30 AM

Angular Momentum BFDBFDBFD FND FN

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