This preview shows pages 1–2. Sign up to view the full content.
Worksheet N: Rigid body motion (II)
1.
Several forces
F
are applied to the 2D objects below. All forces have equal magnitude.
All the objects have the same mass, uniformly distributed.
.
Rank the net torque about the center of mass for each of the objects.
Drawn in red are the leverarm vectors r for each of the forces for rotations about the
center of mass.
In case A, the torque is simply
2
A
d
F
τ
=
.
In case B, it’s a little larger, because of the additional force (note that the new
torque is less than Fd/2 because r and F are not perpendicular). So
AB
<
In case C, the vertical force is applied on the center of mass, and thus r = 0; the
horizontal force is parallel to the corresponding r and thus r×F = 0. Thus
0
C
=
In case D, the force that is directed toward the center of the disk is antiparallel to
the corresponding r, thus r×F = 0. The other two forces produce exactly the same
torque as in case B (assuming that the angle is the same, as suggested by the
figure). Thus
D
B
=
In conclusion,
0
DBAC
τττ
=>>=
d
d
d
d
/2
[A]
[B]
[C]
[D]
r
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 HerreraSiklody
 Force, Mass, Work

Click to edit the document details