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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 cm3 of an ideal gas at 273K and 105 Pa is 2.7x1019.
What is the number of molecules in 1 cm3 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) = v2/L
(c) the tension is the weight of the bob (Mg) plus the force in the string due to the centripetal force (= Mv2/L)
Total tension = Mg + Mv2/L

2. PV = NkT
where N is the number of molecules and k is Boltzmann's constant
P1 = 105 Pa P2 = 10-4 Pa N1 = 2.7x1019.
T1 = T2 = 273 K = T V1 = V2 = 1 cm3 = V
P1/N1 = kT/V = P2/N2

Therefore: N2 = N1[P2/P1] = 2.7x1010 molecules per cubic cm
 

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