Understanding Angular Momentum and Torque: Conservation & Vector Product, Study notes of Acting

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Conservation Theorems: Angular Momentum

Vector Nature of Rotation

The torque is expressed mathematically as a vector product of r and F

τ = r x F

If F and r are both perpendicular to the z axis

τ is parallel to the z axis

Vector product (cont’d)

The direction of A x B is given by the right-hand rule when the fingers

are rotated from the direction of A toward B through an angle ø

This defines a right-handed cartesian system

Vector product (cont’d)

If we take the vector product by going around the figure in the direction of the arrows (clockwise) the sign is positive

i x j = k

Going around against the arrows the sign is negative

i x k = -j

^ ^ ^

Throughout this course we adopt right handed coordinate systems

Angular momentum (cont’d)

The figure shows a particle of mass m attached to a circular disk of negligible mass moving in a circle in the xy plane with its center at the origin The disk is spinning about the z-axis with angular speed ω

L = r x p = r x mv = r m v k = m r ²ω k = mr ²ω

The angular momentum is in the same direction as the angular velocity vector. Because mr ² is the moment of inertia for a single particle we have

L = I ω

^ ^

Angular momentum (cont’d)

The angular momentum of this particle about

a general point on the z axis is not parallel

to the angular velocity vector

The angular momentum L’ for the same

particle attached to the same disk but with

L’ computed about a point on the z axis

that is not at the center of the circle

Conservation of angular momentum

If the net external torque acting on a system about some point is zero, the total angular momentum of the system about that point remains constant. ☛ The angular momentum of a particle (with respect to an origin from which the position vector r is measured ) is

L = r x p

☛ The torque (or moment of force) with respect to the same origin is

τ = r x F

Position vector from the origin to the point where the force is applied

τ = r x p

L = ( r x p ) = ( r x p ) + ( r x p )

d dt But of course r x p = r x mv = m ( r x r ) = 0

L = r x p = τ

If τ = 0 L = 0 L is a vector constant in time

A figure skater can increase his spin rotation rate from an initial rate of 1 rev every 2 s to a final rate of 3 rev/s. If his initial moment of inertia was 4.6 kg m ² what is his final moment of inertia? How does he physically accomplish this change?

I = 0.77 kg m ²

f

(a) What is the angular momentum of a figure skater spinning at 3.5 rev/s with arms in close to her body, assuming her to be a uniform cylinder with height of 1.5 m, radius of 15 cm, and mass of 55 kg? (b) How much torque is required to slow her to a stop in 5 s, assuming she does not moves her arms.

L = 14 kg m ² /s

τ = -2.7 mN

Jerry running away from Tom jumps on the outside edge of a freely turning ceiling fan of moment of inertia I and radius R. If m is the mass of Jerry, by what ratio does the angular velocity change?

ω I + m R ²

I

Angular Momentum of a System of Particles

Newton’s second law for angular motion The net external torque about a fixed point acting on a system equals the rate of change of the angular momentum of the system about the same point

net, ext

dL

sys

dt

Angular impulse

∆ L = ∫ τ dt

tf

ti

sys net, ext It is often useful to split the total angular momentum of a system about an arbitrary point O into orbital angular momentum and spin angular momentum

L = L + L

orbit spin sys

Angular Momentum of a System of Particles Newton’s second law for angular motion (cont’d) Earth has spin angular momentum due to its spinning motion about its rotational axis and it has orbital angular momentum about the center of the Sun due to its orbital motion around the Sun Assuming the Earth is a uniform sphere L = I ω = M R ² ω = 7.1 x 10 kg m ² /s

spin dally^ dally

L = rcm x M v = M r ² ω = 2.7 x 10 kg m ² /s cm cm orbit yearly

Pulling Thgrough a Hole (cont’d)

Because the particle is being pulled in slowly yhe acceleration is virtually the same as if the particle were moving in a circle

Pulling Thgrough a Hole (cont’d)