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Physics 8
Spring 2012
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Quiz 4 - Solutions
Make sure your name is on your quiz, and please box your final answer. Because we will
be giving partial credit, be sure to attempt all the problems, even if you don’t finish them!
A mass m1 undergoes circular motion of
radius R on a horizontal frictionless table,
connected by a massless string through a
hole in the table to a second mass m2 . If
m2 is stationary, find expressions for
1. the string tension,
2. the velocity of m1 , and
3. the period of the circular motion (express your answer in terms of the
masses, m1 and m2 , g, and the radius
of the circle, R, and not the velocity,
v of m1 ).
————————————————————————————————————
Solution
1. The tension in the string comes from the hanging mass. The spinning mass provides
support against m2 falling any further down, but that’s simply the origin of the tension.
Looking at the hanging mass, m2 , the only forces acting on it are the weight downward,
−m2 g, and the tension, T , up. Balancing the two gives
T = m2 g.
2. The weight of the hanging mass, m2 , provides the centripetal force that the spinning
mass, m1 , needs to keep it from sliding off the table due to its inertia. So, the weight
from the hanging mass is m2 g, while the centripetal force is m1 v 2 /R, where v is the
velocity of m1 . Setting the two equal gives
m1 v 2
= m2 g.
R
Thus, the velocity is
r
v=
1
m2
gR.
m1
3. The period of the motion is T , and is simply the distance that m1 goes (the circumference, 2πR), divided by how fast it goes around (v found above). So, T = 2πR/v.
Plugging in the values we found in question 2 above we find
s
r
2πR
m1
m1 R
= 2π
.
= 2πR
T =
v
m2 gR
m2 g
2