Preparation problems
6-24. Suppose the moon were held in orbit not by
gravity but by the tension in a massless cable. Estimate the magnitude
of the tension in the cable.
Solution:
Were the moon held in orbit by a massless cable, the tension
in the wire would be sufficient to provide the force necessary to keep
the moon in a circular orbit. In other words,
So to find the tension in the cable, we need to know the radial
acceleration of the moon. This we find using our uniform circular
motion equation from our constants and equation sheet:
Recalling that r is the distance between the Earth and the moon, we
can look up the values for mmoon, r and v from Appendix E:
Putting this all together, we get
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(7.35 ×1022 kg) |
(1000 m/s)2
3.85 ×108 m
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6-26. 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 (see Figure 6-59 on
page 146). If m2 is stationary, find (a) the tension in the
string and (b) the period of the circular motion.
Solution:
The circular motion of the first mass is the result of the
force being applied to the mass by the string. We can find the tension
in the string by looking at a free-body diagram for the objects.
(a) Since we know the second mass is not moving, and
hence not accelerating, we know the sum of the forces acting on it must
be zero. This means we can quickly write down
(b) Now we need to find the period of rotation for the
first mass. We can see that the forces in the y-direction balance, as
the mass is remaining on top of the table. Thus the net force in the
x-direction must be responsible for the circular motion. This means
Since we know that the radial acceleration is given by
where t is the period of rotation, we can substitute into our
equation above to get
and we are done.
6-69. A space station is in the shape of a hollow ring
with an outer diameter of 150 m. How fast should it rotate to simulate
Earth's surface gravity-that is, so that the force exerted on an object
by the outside wall is equal to the object's weight on Earth's surface?
Solution:
We want a situation in which the normal force exerted by the
wall is equal to the mass of an object times its radial acceleration, or
But on the surface of the earth, this normal force would be equal to the
object's weight, or
So we quickly see that wee need the station to rotate such that
Since we know that ar = v2/r, we see that
Since we want to know how fast the station spins, we should solve for
the period, T. This gives us
The station rotates one time in 17.4 s, so its angular velocity is
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w = |
1
17.4 s
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= 0.058 rev/s = 3.45 rev/min |
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Solutions translated from TEX by TTH, version 1.57.
