Sunday, June 10, 2007

Force, Momentum and Circular Motion

You might have noted that the angle between the force acting on a body and the momentum of the body can be anything between zero and 2π. (But the change of momentum is always in the direction of force). Here is a simple question which you should be able to answer in a minute. If you find it difficult you should work harder to understand basic points in mechanics thoroughly.

A particle moves in a plane such that the rectangular components of its momentum vary simple harmonically with the same period and amplitude, but with a constant phase difference of π/2. The angle (in radian) between the momentum of the particle and the force acting on it is

(a) π (b) π (c) zero (d) π/2 (e) varying between zero and 2π

The particle is forced to move simple harmonically along two mutually perpendicular directions. In other words, this is a case of the superposition of two simple harmonic motions of the same perod and amplitude at right angles to each other. Since the phase difference is π/2, the resultant motion is uniform circular motion. The angle between the resultant momentum of the particle and the force acting on the particle is therefore π/2.

Now consider the following MCQ which appeared in IIT-JEE 2007 question paper:

A particle moves in the X-Y plane under the influence of a force such that its linear momentum is p(t) = A[î cos(kt) – ĵ sin(kt)], where A and k are constants. The angle between the force and the momentum is

(a) 0º (b) 30º (c) 45º (d) 90º

Simply by noting that the momentum vector p has simple harmonically varying components A cos(kt) and A sin(kt) in the X and Y directions respectively, you can conclude that this is a case of the superposition two simple harmonic motions of the same frequency and amplitude at right angles, with a constant phase difference of π/2. [The phase difference is π/2 since one is a sine function while the other is a cosine function].

So, this is uniform circular motion and the angle between momentum and force, as usual, is π/2.

Now, if you want to give a quantitative proof, you may proceed as follows:

The time rate of change of momentum is force. Therefore, force

F = dp/dt = –Ak[î sin(kt) + ĵ cos(kt)]

The angle θ between the force vector F and the momentum vector p is given by

Cosθ = F.p/ Fp, where the bold face letters represent vectors.

But F.p = –Ak[î sin(kt) + ĵ cos(kt)] . A[î cos(kt) – ĵ sin(kt)]

= –A2k sin(kt)cos(kt) + A2k cos(kt)sin(kt) = 0

Since cosθ = 0, θ = π/2.

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