Chapter 11: Rolling, Torque, and Angular Momentum

Lecture 2310/21/2009

RotationGoals for this Lecture:

Study the rolling of circular objects and its relationship with friction

Redefine torque as a vector to describe complex rotational problems

Introduce the concept of Angular Momentum of single particles and systems or particles

Newton’s second law for rotational motionConservation of angular momentumApplications of the conservation of angular momentum

TorqueIf a particle has a position given by the vector r, is acted on by a force F, it experiences a torque given by:

! = r x F

B

Torque Dropsa)You look directly up to see a rain drop 10m

above your head falling directly at you. Ignoring the drag force, what is the torque that YOU observe to be acting on this drop?

b)There is another raindrop that is 10m above your friend’s head, who is 5m away from you. Ignoring the drag force, what is the torque that YOU observe to be acting on this drop?

Torque Drops

x

y

Fg

r1

a) !1 = r1 x F = r mg sin(180°) = 0

Fg = 0î + -mg " +0k !

r2 = rxî + ry" + 0k !

""""= 5î + 10" + 0k !

b) !2 = r2 x F = |r| mg sin(#)

or = 0î + 0" + 5 mg k ! ! 0

Same force, acting on the “same” object produces different torque. The

reference point matters!!

Angular MomentumFor an object moving with respect to an origin O, we define its angular momentum as

l = r x p = r x (mv) = m(r x v) = mr!v = mrvsin#

Units: l = mrvsin# = [kg][m][m/s] = [kg m2/s] = [kg m2/s2·s] = [J·s]

Newton’s Second LawWe can now write the equivalent expression of Newton’s second law for rotation:

Derivation of this expression:

Translation

Fnet = dp/dt

Rotation

!net = dl/dt

d!l

dt=

d(!r ! !p)dt

= md(!r ! !v)

dt= m

!d!r

dt! !v + !r ! d!v

dt

"= m (!v ! !v + !r ! !a)

= m(0 + !r ! !a) = !r ! (m!a) = !r ! !Fnet = !r !#

!Fi =#

!r ! !Fi =#

!"i = !"net

Water balloonYou lean over the edge of Hume’s roof and drop a water balloon. For someone on the ground, 10m away from the bldg what is a) the angular momentum they observe for the

balloon as a function of time?

b) What is the torque they observe exerted on the balloon by gravity?

c) Is the relationship !net = dl/dt still true?

10m

20m

Water balloona) the angular momentum they observe for the

balloon as a function of time?

L = r x p = r(t) x mv(t) = m r(t) v(t) sin#(t) = m r!(t) v(t), r! = 10 m

= m r! v(t) = m r! [v0 + gt] = (m r! gt)k!

L increases linearly with time.

10m

20mr#

Water balloonb) What is the torque they observe exerted on

the balloon by gravity?

! = r x F = r(t) Fg sin#(t) = r!(t) Fg, r! = 10 m

= mgr! k ! = constant

c) Is the relationship !net = dL/dt still true?

l = (m r! gt)k! -> dl/dt = mr!g k ! = !

10m

20mr#

Quiz #

Consider two objects being twirled (uniform circular motion) around a hand by two ropes of different lengths r1 > r2. The tension in both ropes are equal. What is the ratio of the angular momentum between the two (L1/L2)?

a) L1/L2 > 1

b) L1/L2 = 1

c) L1/L2 < 1

d) There is not enough information to determine an answer

T#

Two particles are moving in a plane as shown in the figure. Particle 1 has mass 10 kg and velocity 2.0 m/s î. Particle 2 has mass 3.0 kg and velocity 3.0 m/s "a) What is net angular momentum of the two

particles about the origin O (both magnitude and direction)?

Net Angular Momentum

x

y

d 1 =

1.5

m

d2 = 3.0m

O

a) What is net angular momentum of the two particles about the origin O (both magnitude and direction) ?

l1 = r1xp1 = (d1") x (m1v1î) = d1m1v1 (-k !) = -d1m1v1 k !

l2 = r2xp2 = (d2î) x (m2v2") = d2m2v2 (+k !) = +d2m2v2 k !Ltot = l1 + l2 = (d2m2v2 - d1m1v1)k!

= (27 - 30)k !

= -3 k !

Net Angular Momentum

x

y

d 1 =

1.5

m

d2 = 2.8m

O