Physics 101
Classical Physics
Spring 2010
Homework 7 Solutions
DUE IN CLASS ON March 18
1. Scott and Tina are playing on a seesaw which is 4 m long and has a fulcrum in
the middle. Tina is 30 kg and sits at one end, while Scott who is 40 kg sits so
that they balance. Assume the seesaw by itself is uniform and balanced. How
far from the end should Scott sit in order to achieve balance?
(a) 1 m,
(b) 0.5 m
, (c) 3 m, (d) 1.5 m, (e) 0.75 m
If the seesaw is balanced, there is zero angular acceleration, so the sum of the torques should
be zero. With the origin at the pivot point,
0 =
X
~
τ
=
m
s
r
s

m
t
r
t
⇒
r
s
=
m
t
m
s
r
t
=
30 kg
40 kg
(2 m) = 1
.
5 m
This is the distance from the pivot point, so Scott should sit 2 m1.5 m=0.5 m from the end.
2. If the net torque acting upon a system is zero, which of the following is true?
(a) The system can have no translational motion
(b) The system can have no rotational motion
(c) The system can have neither rotational or translational motion
(d) The system may be rotating if and only if it is rotating at a constant speed
about its center of gravity
(e) The system may be rotating about any point if and only if it is rotating at
a constant speed
The system can have both translational and rotational motion with a net torque of zero,
however, the rotational motion must be at a constant angular velocity about the center of
mass.
If the system were rotating about a point away from the center of mass, then the
center of mass would be rotating. If the center of mass is rotating, there must be a net torque
to provide the accleration.
Therefore, if there is zero net torque the rotation at constant
speed must be about the center of mass.
3. Imagine a sphere, a disk, and a hoop that all have the same mass and radius. If
the three are raced down an incline plane, which of the following describes the
order in which they reach the bottom, from first to last?
(a) disk, sphere, hoop
(b) sphere, disk, hoop
(c) hoop, disk, sphere
(d) hoop, sphere, disk
(e) they all reach the bottom at the same time
They will reach the bottom in order of increasing moment of inertia. Since the hoop has all
of its mass at the radius, it will have the largest moment of inertia. The sphere has more
of its mass close to the point of rotation relative to the disk, so the sphere will have the
smallest moment of inertia. Therefore, the sphere finishes first followed by the disk followed
by the hoop.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
4. A uniform plank XY (where X and Y denote the ends of the plank) is supported
by two equal upward 120N forces at X and Y. The support at X is then moved
towards Y to a point Z located a quarter of the plank’s length away from point
X. The forces applied at Y and Z are now those needed to support the plank.
The new supporting force at Y is:
(a) 40 N,
(b) 80 N
, (c) 60 N, (d) 240 N, (e) 160 N
With the supports at X and Y, there was a force of 120 N at each point, so the weight of
the plank is
mg
=240 N. If we take the point Y to be the pivot point, the force due to the
gravity provides a torque at
L/
2, and the support at point Z provides a torque at 3
L/
4. If
we let
F
z
be the force at
Z
, then
0 =
X
~
r
×
~
F
=

This is the end of the preview.
Sign up
to
access the rest of the document.