- #1
gnits
- 137
- 46
- 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.
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.