To find the relative velocities of linear and circular motion

[gnits]

Homework Statement

To find relative velocities

Relevant Equations

d=st

Could I please ask for help regarding the final part of the following question:

It is the very last part, to find v in terms of u.

So I have that the velocity of the midpoint of XY is:

V_m = (u/2) i + (u/2) j

I let the position vector of P be:

r_p = cos(wt) i + sin(wt) j

(w = angular velocity = v because v = rw and r = 1. So I am assuming that the particle is at (1,0) when t = 0)

Differentiating this gives the velocity of P as:

V_p = -w sin(wt) i + w cos(wt) j

And so the velocity of P relative to M is:

V_p - V_m = V_pm = (-w sin(wt) - u/2) i + (w cos(wt) - u/2) j

and so, the modulus of V_pm is given by:

|V_pm|^2 = w^2 (sin(wt))^2 + uw sin(wt) + u^2/4 + w^2 (cos(wt))^2 - uw cos(wt) + u^2/4

Which simplifies to:

|V_pm|^2 = w^2 + u^2/2 + uw (sin(wt) - cos(wt) )

This will be maximal when cos(wt) = 0 and sin(wt) = 1

Simplifying and substituting v for w, and replacing |V_pm| with u leads to:

v^2 + u^2/2 + uv = u^2

This does not lead to the provided answer of v = u - u/sqrt(2)

Thanks for any help,
Mitch.

 

[PeroK]

Homework Statement:: To find relative velocities
Relevant Equations:: d=st

Could I please ask for help regarding the final part of the following question:

View attachment 264411

Which simplifies to:

|V_pm|^2 = w^2 + u^2/2 + uw (sin(wt) - cos(wt) )

This will be maximal when cos(wt) = 0 and sin(wt) = 1

This is the mistake.

 

[gnits]

This is the mistake.

Thanks for your reply, it helped me see my error.
I was indeed wrong to say that the max value will occur when cos(wt) = 0 and sin(wt) = 1
As I have the sum of three terms and the first two are positive, I will need to maximize the last term.
I wrongly stated this max as 1 but it is in fact sqrt(2)
This leads to the correct answer
Thanks again.