Physics 18 Fall 2009
Midterm 1 Solutions
For the midterm, you may use one sheet of notes with whatever you want to put on it, front
and back. Please sit every other seat, and
please don’t cheat
! If something isn’t clear, please
ask. You may use calculators. All problems are weighted equally.
PLEASE BOX YOUR
FINAL ANSWERS!
You have the full length of the class. If you attach any additional
scratch work, then make sure that your name is on every sheet of your work.
Good luck
!
1. A typical laboratory centrifuge rotates at 4000 RPM. Test tubes have to be placed into
a centrifuge very carefully because of the very large accelerations.
(a) What is the acceleration at the end of a test tube that is 10 cm from the axis of
rotation?
(b) For comparison, what is the magnitude of the acceleration a test tube would
experience if dropped from a height of 1.0 m and stopped in a 1.0 ms (10

3
seconds) encounter with a hard ﬂoor?
————————————————————————————————————
Solution
(a) Since it’s spinning, the test tube is subjected to a centripetal acceleration
a
cent
=
v
2
/r
, where
v
is the rotational velocity and
r
is the rotational distance. Now,
we know the rotational frequency,
f
= 4000 rotations/min
÷
60 sec/min
≈
67
rotations per second. This means that the period is
T
= 1
/f
= 1
/
67 = 0
.
015
seconds. Now, the period is the time it takes to go around one orbit in the
centrifuge, and is the total distance around (2
πr
) divided by how fast it’s going
around (
v
). So,
T
=
2
πr
v
,
which means that
v
= 2
πr/T
. Thus, the acceleration is
a
cent
=
v
2
r
=
(2
π
)
2
T
2
r.
Plugging in the numbers gives
a
cent
=
(2
π
)
2
.
015
2
×
0
.
1 = 1
.
75
×
10
4
m/s
2
, which is
1790
g
!
(b) If we drop a test tube, it experiences a big acceleration when it hits the ground.
Since it goes from some velocity
v
drop
to zero over a time Δ
t
, then the acceleration
is
a
=
v
drop
/
Δ
t
. What’s the initial speed? Since it’s dropped from a height

h
,
under the constant acceleration due to gravity then
v
2
f
=
v
2
i

2
g
(

h
) = 0 + 2
gh
,
since the tube is assumed to have dropped from rest. The ﬁnal velocity it gets
is just
v
f
=
v
drop
=
√
2
gh
. Thus,
a
=
√
2
gh
Δ
t
=
√
2
×
9
.
8
×
1
10

3
= 4400 m/s
2
, which is
about 450
g
. So, the centrifuge subjects the tube to accelerations even greater
than hitting the ground!
1