Angular Momentum and Rotational Motion, Slides of Physics

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More Angular Momentum, then

Statics

Vector Angular Momentum of a Particle

• A particle with momentum is at

position from the origin O.

• Its angular momentum about the

origin is

• This is in line with our definition

for part of a rigid body rotating about an axis: but also works for a particle flying through space.

p

r

L = r × p

Viewing the x -axis as coming out of the slide, this is a “right-handed” set of axes: (^) ˆ ˆ ˆ i × j = + k

p

r^ 

L = r × p

O x

z

y

m

Rotational Motion of a Rigid Body

• For a collection of interacting particles, we’ve seen

that

the vector sum of the applied torques, and the being measured about a fixed origin O.

• A rigid body is equivalent to a set of connected

particles, so the same equation holds.

• It is also true (proof in book) that even if the CM is

accelerating,

/ (^) i i

dL dt = (^) ∑ τ

L

τ i

dL CM (^) / dt = (^) ∑ τ CM

A dumbbell (two small masses at the ends of a

light rigid rod) is mounted on a fixed axle

through its center, at an angle θ. It is set in

steady rotation. The direction of the angular

momentum of the system is:

A. Along the axle

B. Along the dumbbell rod

C. Neither of the above.

Spinning Top

• Pointing your right thumb in the direction of the angular velocity vector , your curling fingers point in the direction of rotation.
• Gravity exerts a torque about the pivot point , evidently directed inwards.
• From will be inwards, the tip of is describing a horizontal circle: this is “precession”. - a

d

mg



ω^ ^ 

τ = d × mg

τ = dL / dt = Id ω / dt

^ ^ 

d ω

ω

Precession Rate

• The horizontal component of the angular velocity vector has length and it precesses around a circle centered above the pivot point.
• The precession angular velocity is written , where θ measures angle around the horizontal circle.
• If in time dt there is precession through d θ,
• so
  • aθ ω

τ = dL / dt = Id ω / dt

^ ^ 

φ

ω sinφ ω sinφ

Ω = d θ / dt

d ω =( ω sinφ (^) ) d θ 1 1 sin sin

d d mgd dt dt I I

θ ω τ ω φ ω φ ω

= = =

Free Body Diagrams

• To apply Newton’s Laws to find how a body moves, we must focus on that body alone and add all the (vector) forces acting on it.
• The diagram showing all the forces on one body (or even part of a body) is called a “free body diagram”—we’ve “freed” the body from the rest of the system, representing everything else just by the forces on this body.
• The net (total) force then goes into (^) Σ F = ma.

Flat Forces?

• If a body in equilibrium is acted on by three

and only three forces, do the force vectors

have to lie in a plane?

A. Yes

B. No

Clicker Question

• A body is in equilibrium. It is acted on by three

forces, lying in a plane.

• Do the lines of action of the three forces all

pass through the same point?

A. Yes

B. No

Three Force Equilibrium

• If a body is in equilibrium when acted on by

three forces, the three forces must lie in the

same plane AND all pass through a common

point. If they don’t, taking a perpendicular

axis through a point where two of them meet,

the third force gives an unbalanced torque

about that point, so the body will have

angular acceleration.

Clicker Question

• What is the approx tension

T in the top string, given the mass is 2 kg, and it’s hung from the midpoint of the rod, which is light and hinged, the angle is 30°?

A. 10 N

B. 20 N

C. 20 √3 N

D. 40 N

• a