11-1
Chapter 11
Questions
The torque for a given force τ=rF sin φ, where φis the angle between ~r and ~
F. In this case(1)
~r is from a given point to the origin, where the forces are applied. The forces all have the same
magnitude, and ris the same for each force, so only the angle φis different for the three forces.
(a) For P1,~
F1is perpendicular to ~r (φ= 90◦), so produces a maximum torque; ~
F3is parallel to ~r
(φ= 0◦), so produces zero torque; and ~
F2is intermediate. Thus the order is F1,F2,F3.
(b) For P2, both ~
F1and ~
F2are perpendicular to ~r, and ~
F3has φ < 90◦. Thus the order is
F1-F2tied, F3.
(c) For P3, both ~
F1and ~
F3are perpendicular to ~r, and ~
F2has φ < 90◦. Thus the order is
F1-F3tied, F2.
We have a particle moving past a fixed point. In this situation, the most useful expression for the(6)
angular momentum is eq. 11-21, `=mr⊥v, where r⊥is the perpendicular distance from the point
to a line along ~v. In this case ~v is along a horizontal line at y= 1. The distance r⊥= 0 for each
point will simply be its ydistance from that line, |∆y|. The values of |∆y|are 0 for a, 3 for b, 2 for
c, 2 for d, and 0 for e. So the order is b,c-dtied, a-etied .
(a) The slab and gum collision will conserve angular momentum. The larger the gum’s angular(10)
momentum (with respect to point O) before the collision, the larger their angular momentum, and
therefore the larger the angular speed, afterward. I think the most convenient expression for the
angular momentum of the gum (a particle) beforehand is eq. 11-21, `=mr⊥v, where r⊥is the
perpendicular distance from Oto a line along ~v. The gum has the same mass and speed for each
path, so only r⊥is different.
Paths 2, 3, and 5 point directly at O, so the gum has zero angular momentum. Looking at the
others, r⊥is 1/2 of the diagonal for path 4, 1/2 of the long side for path 6, 1/2 of the short side for
path 7, and roughly 1/4 of the short side of the slab for path 1. So the order from greatest to least
initial angular momentum and final angular speed is 4, 6, 7, 1, 2-3-5 tied .
(b) By convention, the angular momentum is negative if the rotation is clockwise, which would be
the case for paths 1, 4, and 7 .
Problems
For particle 1, the mass m1= 6.5 kg, the speed v1= 2.2 m/s, and the distance from point O d1=(27)
1.5 m. For particle 2, m2= 3.1 kg, v2= 3.6 m/s, and d2= 2.8 m.
(a) The magnitude of the angular momentum of a particle about a point `=mr⊥v(eq. 11-21). As
shown in the diagram, d1and d2are the perpendicular distances (r⊥), so `=mdv:
`1=m1d1v1= (6.5 kg)(1.5 m)(2.2 m/s) = 21.45 kg·m2/s
and
`2=m2d2v2= (3.1 kg)(2.8 m)(3.6 m/s) = 31.25 kg·m2/s.
From the right-hand rule (pp. 50 and 283), ~
`1is into the page and ~
`2is out of the page. Because
they’re in opposite directions, the net angular momentum will be the difference of the two magni-
tudes: L=`2−`1= 31.25 −21.45 kg·m2/s = 9.8 kg·m2/s .
