MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
Problem Solving Session 2: Circular Motion Dynamics
IC_W04D3-2 Tension in Strings Solution
Two objects 1 and 2 of mass m1 and m2 are whirling around a shaft with a constant
angular velocity  . The first object is a distance d from the central axis, and the second
object is a distance 2d from the axis. You may ignore the mass of the strings and neglect
the effect of gravity.
a) What is the tension in the string between the inner object and the outer object?
b) What is the tension in the string between the shaft and the inner object?
Solution: We begin by drawing separate force diagrams for each object.
Newton’s Second Law, F1  m 1 a1 , for the inner mass in the radial direction is
rφ: T2  T1  m1 d  2 .
Newton’s Second Law, F2  m 2 a 2 , for the outer mass in the radial direction is
rφ:  T2  m2 2d  2 .
From the force equation for outer object, the tension in the string between the inner object
and the outer object is
T2  m2 2d  2 .
Using this result for T2 in the force equation for the inner object yields
m2 2d  2  T1  m1 d  2
which can be solved for the tension in the string between the shaft and the inner object
T1  d  2 (m1  2m2 ) .
IC_W04D3-3 Tension in a Spinning Rope Solution
A uniform rope of mass m and length L is attached to shaft that is rotating at constant
angular velocity  . Find the tension in the rope as a function of distance from the shaft.
You may ignore the effect of gravitation. Hint consider a small segment of the rope and
apply Newton’s Second Law to that segment to find a differential equation that describes
how dT / dr varies with respect to distance r from the shaft.
Solution:
Divide the rope into small pieces of length r , each of mass m  (m / L)r . Consider
the piece located a distance r from the shaft. The radial component of the force on that
piece is the difference between the tensions evaluated at the sides of the piece,
Fr  T (r  r)  T (r) .
The piece is accelerating inward with a radial component ar  r  2 . Thus Newton’s
Second Law becomes
Fr  m 2 r
T (r  r)  T (r)  (m / L)r r 2 .
(0.1)
Denote the difference in the tension by T  T (r  r)  T (r) . After dividing through by
r Eq. (0.1) becomes
T
 (m / L) r 2
r
In the limit as r  0 , Eq. (0.2) becomes a differential equation,
(0.2)
dT
 (m / L) 2 r .
dr
(0.3)
From this, we see immediately that the tension decreases with increasing radius. We
shall solve this equation by integration
rr
T (r)  T ( L) 
r
dT
dr   (m 2 / L)  r  dr 
L
dr 
r L

 (m 2 / 2L)(r 2  L2 )
(0.4)
 (m 2 / 2L)(L2  r 2 ).
We use the fact that the tension, in the ideal case, will vanish at the end of the rope,
r  L . Thus,
T (r)  (m 2 / 2L)(L2  r 2 ).
(0.5)
This last expression shows the expected functional form, in that the tension is largest
closest to the shaft, and vanishes at the end of the rope.
ICW04D3-4 Double Star System
Consider a double star system under the influence of gravitational force between the
stars. Star 1 has mass m1 and star 2 has mass m2 . Assume that each star undergoes
uniform circular motion such that the stars are always a fixed distance s apart (rotating
counterclockwise in the figure below). What is the period of the orbit?
Solution: Because the distance between the two stars doesn’t change as orbit about each
other there is a central point where the line connected the two objects intersect as the
objects move as can be seen in the figure above. (We will see later on in the course that
central point is the center of mass of the system.) Choose radial coordinates for each star
with origin at that central point. Let rö1 be a unit vector at Star 1 pointing radially away
r
from the center of mass. The position of object 1 is then r1  r1 rφ1 , where r1 is the distance
from the central point. Let rö2 be a unit vector at Star 2 pointing radially away from the
r
center of mass. The position of object 2 is then r2  r2 rφ2 , where r2 is the distance from
the central point. Because the distance between the two stars is fixed we have that
s  r1  r2 .
The coordinate system is shown in the figure below.
Coordinate system for double star orbits
The gravitational force on object 1 is then
v
Gm1m2
F2,1  
rφ1 .
s2
The gravitational force on object 2 is then
v
Gm1m2
F1,2  
rφ2 .
s2
The force diagrams on the two stars are shown in the figure below.
Force diagrams on objects 1 and 2
Let  denote the magnitude of the angular velocity of each star about the central point.
Then Newton’s Second Law, F1  m 1 a1 , for Star 1 in the radial direction rφ1 is
G
We can solve this for r1 ,
m1 m2
s
2
 m1 r1  2 .
r1  G
m2
.
 2s2
Newton’s Second Law, F2  m 2 a 2 , for Star 2 in the radial direction rφ2 is
G
m1 m2
s
2
 m2 r2  2 .
We can solve this for r2 ,
r2  G
m1
.
 2s2
Since s , the distance between the stars, is constant
s  r1  r2  G
m2
m
m  m 
 G 21 2  G 2 2 2 1
2 2
.
 s
 s
 s
Thus the magnitude of the angular velocity is
 m  m  
  G 2 3 1 
s


12
and the period is then
 4 2s3 
T


  G m2  m1 
2


12
(0.6)
.
Note that both masses appear in the above expression for the period unlike the expression
for Kepler’s Law for circular orbits.