Physics 8 — Wednesday, November 2, 2011

Monday’s popular vote was for Dec 1 (Thu) at 7pm to meet withRichard Farley about the connection between physics andarchitecture / structures. Normal homework study session will startjust after this discussion wraps up.

I made a mistake in solving problem 2 on homework 6. As nearlyeveryone figured out, the correct answer was 1.73 hours, not the1.0 hours that I got. The problem is that my solution used (0,1)instead of (1,0) for “east” on the first step.

I will be at this Thursday’s 7pm homework session. Zoey will dotonight’s session.

Recap from Monday / last week’s reading, this week’s HWFor motion in a circle, acceleration has a centripetal componentthat is perpendicular to velocity and points toward the center ofrotation. If we put the center of rotation at the origin (0, 0) then

x = R cos θ y = R sin θ

~r = (R cos θ,R sin θ) = R (cos θ, sin θ)

The “angular velocity” ω is the rate of change of the angle θ

ω =dθdt

The units for ω are just s−1 (which is the same as radians/second,since radians are dimensionless). Revolutions per second are ω/2π,and the period (how long it takes to go around the circle) is 2π/ω.The velocity is

~v =d~rdt

= ωR (− sin θ, cos θ), |~v | = ωR

The magnitude of the centripetal acceleration (required rate ofchange of ~v , to keep object on circular path) is

ac = ω2R =v2

R

and the centripetal force (directed toward center of rotation) is

|~Fc | = mac = mω2R =mv2

R

Moving in circle at constant speed (velocity changes, speed doesnot) is called uniform circular motion. For UCM, ~a ⊥ ~v , andω = constant. Then

~a =d~vdt

= −ω2R (cos θ, sin θ) = −v2

R(cos θ, sin θ)

(For non-UCM case where speed is not constant, ~a has anadditional component that is parallel to ~v .)

We can also consider circular motion with non-constant speed, justas we considered linear motion with non-constant speed. Then weintroduce the angular acceleration

α =dωdt

=d2θ

dt2

and we can derive results that look familiar but with substitutions

x → θ, v → ω, a→ α, m→ I , p → L

if α is constant (which is a common case for constant torque),then:

θf = θi + ωt +1

2αt2

ωf = ωi + αt

ω2f = ω2

i + 2α (θf − θi )

(If you use these equations, be careful about cases where θ wrapsaround at 2π, or you can get an answer that makes no sense.)

Rotational inertia (“moment of inertia”)

I =∑

mr2 →∫

r2dm

Rotational inertia (“moment of inertia”)

I =∑

mr2 →∫

r2dm

Parallel axis theorem

If an object revolves about an axis that does not pass through theobject’s center of mass (suppose axis has ⊥ distance ` fromc.o.m.), the rotational inertia is larger, because the object’s c.o.m.revolves around a circle of radius ` and in addition the objectrotates about its own center of mass. This larger rotational inertiais given by the parallel axis theorem:

I = Icm + M`2

where Icm is the object’s rotational inertia about an axis (whichmust be parallel to the new axis of rotation) that passes throughthe object’s c.o.m.

I twirl a basketball of mass M, radius R, on a string of length `.The length ` is measured from my fingertips to the surface of thebasketball, where the string is taped down. What is the rotationalinertia for twirling the ball about my fingertips?

(A) I = 23MR2

(B) I = M`2

(C) I = 23MR2 + M`2

(D) I = 23MR2 + M(`+ R)2

(E) I = M(`+ R)2

Kinetic energy has both translational and rotational parts:

K =1

2mv2

cm +1

2Iω2

Angular momentum:

L = Iω = m v⊥ r⊥

(where ⊥ means the component that does not point toward the“reference” axis — which usually is the rotation axis)

HW7, problem 11In the figure below, two identical pucks B and C, each of inertia m,are connected by a rod of negligible inertia that is free to rotateabout its center. Then puck A, of inertia m/2, comes in and hitsB. After the collision, in which no energy is dissipated, what are(a) the rotational speed of the dumbbell, and (b) the linearvelocity of A?

What is the rotational inertia of the B+C system about its axis ofrotation?

(A) I = 12mR2

(B) I = mR2

(C) I = 2mR2

(D) I = 4mR2

Which equation correctlyrepresents conservation ofangular momentum when puckA hits puck B?

(A)mviR = mvAf R + mR2ωf

(B) (m

2

)viR =

(m

2

)vAf R + 2mR2ωf

(C)mviR = mvAf R + 2mR2ωf

Which equation correctlyrepresents conservation ofenergy when puck A hits puckB?

(A)1

2

(m

2

)v2i =

1

2

(m

2

)v2Af +

1

2(2mR2) ω2

f

(B)1

2mv2

i =1

2mv2

Af +1

2(mR2) ω2

f

(C)1

2mv2

i =1

2mv2

Af +1

2(2mR2) ω2

f

Conservation of energy implies (2nd term on right is 12 Iω2):

1

2

(m

2

)v2i =

1

2

(m

2

)v2Af +

1

2(2mR2)ω2

f

Conservation of angular momentum implies:(m

2

)viR =

(m

2

)vAf R + 2mR2ωf

⇒ m

2(vi − vf )R = 2mR2ωf

⇒ vi − vAf = 4Rωf

But this is a hassle to solve, so let’s try another approach. Treatproblem as m

2 colliding elastically with 2m. Momentumconservation implies:(m

2

)vi =

(m

2

)vAf + (2m)vBf

Elastic collision just negates relative velocity:

vBf − vAf = vi ⇒ vBf = vi + vAf

Solving 1st equation (and plugging in vBf from 2nd):

vi = vAf + 4vBf = vAf + 4(vi + vAf ) = 4vi + 5vAf

vAf = −3

5vi

Use 2nd equation again to get vBf :

vBf = vi + vAf = vi −3

5vi = +

2

5vi

So the angular velocity after collision is

ωf = +2

5vi/R

Now let’s check that this solution satisfies the two originalequations written before we reformulated the problem:

vi − vAf = 4vBf = 4Rωf

so angular momentum is conserved. Now energy:

1

2

(m

2

)v2Af +

1

2(2mR2)ω2

f =1

2

(m

2

)(−3

5vi

)2

+1

2(2mR2)

(2

5

vi

R

)2

=m

4· 9

25v2i + m · 4

25v2i =

(m

4

)v2i =

1

2

(m

2

)v2i

so the energy-conservation equation is also satisfied.

For the homework, you can just write down the answer and showthat the answer conserves both angular momentum and energy.

HW7, problem 12

Two skaters skate toward each other, each moving at 3.3 m/s.Their lines of motion are separated by a perpendicular distance of2.0 m. Just as they pass each other (still 2.0 m apart), they linkhands and spin about their common center of mass. What is therotational speed of the couple about the center of mass? Treateach skater as a point particle, one with an inertia of 75 kg andthe other with an inertia of 48 kg.

About what point will the two skaters rotate once they link hands?

(A) About the point halfway between the two skaters

(B) About the center of mass of the two-skater system

(C) About the center of the more massive skater

Iω = m1 v1⊥ r1⊥ + m2 v2⊥ r2⊥

Let’s put skater 1 at x = +1.0 m, with m1 = 75 kg, and skater 2at x = −1.0 m, with m2 = 48 kg. Center of mass is

xcm =(75 kg)(+1.0 m) + (48 kg)(−1.0 m)

75 kg + 48 kg= +0.22 m

So perpendicular displacements from center of mass arer1⊥ = +0.78 m and r2⊥ = −1.22 m. Conserving angularmomentum in “collision” implies:

(m1r21 + m2r

22 )ω = m1v1r1 + m2v2r2

ω =m1v1r1 + m2v2r2

m1r21 + m2r2

2

ω =(75 kg)(3.3 m/s)(0.78 m) + (48 kg)(−3.3 m/s)(−1.22 m)

(75 kg)(0.78 m)2 + (48 kg)(−1.22 m)2= 3.3/s

If you look at the face of a clock, whose hands are movingclockwise, do the rotational velocity vectors of the clock’s handspoint toward you or toward the clock?

(A) Toward me

(B) Toward the clock

(C) Neither — when I curl the fingers of my right hand toward theclock, my thumb points to the left, in the 9 o’clock direction

When you are about to fall over the edge of a precipice, youinstinctively wheel your fully extended arms rapidly in verticalcircles. What does this accomplish?

(A) It accomplishes nothing because there isno external torque, since my arms are partof my body

(B) When I spin my arms counterclockwise (asdrawn in picture), conservation of angularmomentum pushes my body’s center ofmass clockwise, with my feet as the pivotpoint, and this may help me not to fall

(C) I think argument (B) sounds good intheory, but I’m skeptical about how well itreally works in practice

I want to tighten a bolt to a torque of 1.0 newton-meter, but Idon’t have a torque wrench. I do have an ordinary wrench, a ruler,and a 1.0 kg mass tied to a string. How can I apply the correcttorque to the bolt?

(A) Orient the wrench horizontally and hang the mass at adistance 0.1 m from the axis of the bolt

(B) Orient the wrench horizontally and hang the mass at adistance 1.0 m from the axis of the bolt

If the wrench is at 45◦ w.r.t. horizontal, will the 1.0 kg masssuspended at a distance 0.1 m along the wrench still exert a torqueof 1.0 newton-meter on the bolt?

(A) Yes. The force of gravity has not changed, and the distancehas not changed.

(B) No. The torque is now smaller — about 0.71 newton-meter.

(C) No. The torque is now larger — about 1.4 newton-meter.

The angular equivalent of F = ma

τ = I α

Let’s try it . . .