Monday, July 16, 2007

Irodov Problem 1.19

a) the average speed is simply distance covered (half the circumference) over time given by .
b) the average velocity vector is displacement (the diameter) over time given by .
c) If the constant acceleration is w and initially the particle started with 0 speed then,

The final velocity of the particle at the end of the journey will be . This means the average acceleration during this time will be w given by the equation above.


Anurag Somani said...

Err, I don't think you can assume that the particle started with 0 velocity. nowhere is that mentioned. It is possible that the particle had been moving with that velocity all along and we are considering one (technically half) of it's infinite revolutions.

Anurag Somani said...

Thanks. :)

Krishna Kant Chintalapudi said...

You are right. The answer holds correct only for two cases i) constant acceleration from 0 or ii) constant angular velocity.

In all other cases the acceleration vector can be something else. In fact in general it is (v1-v2)/tau where v1 and v2 are the velocity vectors in the two ends.

Consider for example that the particle starts at a 0 speed. Accelerates at a rate (4PiR/tau^2) for quarter circle - this takes tau/2 seconds. Then it decelerates at the same rate for tau/2 secs more until it reaches half circle point. So the particle covers half circle in tau seconds.

However, the average acceleration is 0 in this case since v1=v2=0.

Unfortunately, Irodov does not specify much in the problem :(, so I was not sure what to do.