Pendulum and molecules
1. A pendulum bob
of mass M hangs from the end of a light string of length L. The bob is pulled to one side until
the string is horizontal and then released.
Find (a) the velocity of the bob, (b) its
acceleration and (c) tension in the string when the string is vertical.
2. The number
of molecules in 1 cm
3 of an ideal gas at 273K and 10
5 Pa is
2.7x10
19.
What is the number of molecules in 1 cm
3 of the gas at 273K
and 10
-4 Pa?
Answers:
1. We assume that the bob is undergoing
simple harmonic motion.
(a) Velocity (v ) = angular velocity x L
(b) Acceleration (a) =
v
2/L
(c) the tension is the weight of the bob (Mg) plus the force in the string due
to the centripetal force (= Mv
2/L)
Total tension = Mg + Mv
2/L
2.
PV = NkT
where N is the number of molecules and k is Boltzmann's constant
P
1 = 10
5 Pa P
2 = 10
-4 Pa N
1 =
2.7x10
19.
T
1 = T
2 = 273 K = T V
1 = V
2 =
1 cm
3 = V
P
1/N
1 = kT/V =
P
2/N
2Therefore: N
2 = N
1[P
2/P
1] =
2.7x10
10 molecules per cubic cm
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