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Question 1 (10
Points)
1D
Collision and Impulse
A uniform rope of mass m
per unit length
hangs vertically from a hook so that the lower
end of the rope just touches the horizontal table as shown in Fig. 1(a). The rope is then
released from the hook and descends onto the table. The rope is assumed to be
completely flexible and when it reaches the table it stops instantaneously. When a length
y of the rope has fallen (see Fig. 1 (b)) what is the force exerted by the rope on the table
in terms of y, m and g?
Fig. 1
Solution
The rope descends via free fall onto the table. As it starts from zero initial velocity its
velocity v is related to the height y it has fallen by: v(y)=[2gy]
1/2
. The length of the rope
which lands on the table during the time dt at that instant is: v(y)dt and the corresponding
mass is mvdt
Based on the assumption that the rope stops instantaneously after it lands on the table, the
rate at which the rope transfers momentum to the table is:
dp/dt=[mvdt]v/dt=mv
2
=m(2gy). According to Newton’s equation of motion this is the
force exerted on the table by the falling rope. However, there is also rope of length my
already on the table and the force they exert due to gravity on the table is myg.
Thus the total force exerted by the rope on the table is 2mgy+mgy=3mgy.
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View Full DocumentQuestion 2 (10 Points)
Energy & Work
Figure 2 shows a massive spring whose total mass is M
s
and whose length when
unstretched is L. A smaller mass m is attached to the end of the spring. The mass m and
the spring both move on a
frictionless
table. The spring, when stretched, behaves like a
rod with Young’s modulus E. When the end of
the spring is stretched by an amount D the
extension of a small element of the spring
around the point x can be assumed to be given
by (x/L)D. Similarly when the mass m at the
end of the spring is moving with velocity V a
point x along the spring can be assumed to
move with velocity (x/L)V.
Fig.2
(a) Calculate the potential energy of
the
mass m plus the spring
when m is displaced by
D.
(b) Calculate the kinetic energy of the mass
m plus the spring
when m is moving with V.
(c) What is the frequency of oscillation of m?
Solution
(a) For a rod of length L, area of cross section A and Young’s modulus E the stress
(=Force F/Area A)=E(strain) where strain is defined as
∆
L/L. Consider the point x along
the spring. When it is extended by amount
∆
x the restoring force on it is: F=EA(
∆
x/x).
∆
x is given by:
∆
x = (x/L)D so : F=EA(D/L). The direction of F is in the negative x
direction. Notice that this force F is independent of x just like the tension in a string is the
same everywhere. Thus the spring constant k of the massive spring is =EA/L. The work
done by external force in stretching the end of the spring by the amount D is then given
by the integral of Fd
l
(where d
l
is the displacement of the end of the spring) from
l
=0 to
D, just like a regular massless spring. Thus the potential energy stored in the spring is :
(1/2)kD
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