Physics Problems
from Boris Korsunsky
Mr. Korsunsky, who formerly taught
at Northfield-Mt. Hermon School and who is now at the Harvard School of
Education (korsunbo@gse.hardvard.edu), has published a supplement to Serway’s
Principles of Physics (Saunders ISBN 0-03-005922-4) titled Challenging
Problems for Physics, which is to say the least aptly titled. In the
October and November 2001 issues of The Physics Teacher he presented
some challenge problems, three of which plus solutions are as follows.
Solutions and problem suggestions are welcomed from students or teachers.
His address is 444 Wellesley St., Weston, MA 02493-2631.
| Problem
1-
A long uniform rope is hung over a light, frictionless pulley as shown.
The left part of the rope rests on a table of height H and the right part
on the floor. Soon after the system is released the rope reaches a constant
velocity v. Find that velocity.
The constant velocity occurs when the weight of the rope on the right
side is balance by the weight on the left side plus the force needed to
lift a "small" section of the rope off the table, or
WB =
WA + 
If l
is the linear density of the rope WB = lg(x
+ H) and |
|
WA = lgx.
Also, Dp/Dt
can be written as v(Dm/Dt)
= v(lDx/Dt)
= lv2
since Dx/Dt
= v .
Making these substitutions,
lg(x +
H) = lgx
+ lv2.
Simplifying and solving, v = 
.
If you think about the dimensions or units, this should not be a
surprise!
| Problem
2-
A thin-walled cylinder of radius R is rotating clockwise about its axis
and is placed in a corner as shown. The kinetic coefficient of friction
is m
between the cylinder and both surfaces. Find the angular acceleration a.
In the figure at right A and B are
the normal forces exerted on the cylinder by the floor and wall, respectively,
while the friction forces f1 and f2
are both opposing the rotation. Apply Newton's first law vertically and
horizontally, |
|
A + mB
= mg
B = mA
Solving simultaneously, 
and 
.
Then the friction forces are given
by 
and 
.
Lastly, applying the second law for
rotation, torque t
= Ia,
with I = mR2 and seeing
both torques as negative,
.
Solving:
