I calculated the polar equation to be r = √(a2cos2(wt) + b2sin2(wt)) r^
and then I differentiated it to get the velocity as v = ((b2 - a2)sin(wt)cos(wt)w)/√(a2cos2(wt) + b2sin2(wt)) r^ + √(a2cos2(wt) + b2sin2(wt))w θ^
The angle does not match. tanθ = b/a√2 is given
Also this is one of the part of the question, and I still can't understand how to wite the polar equation for this curve
1. The question
The position of a particle is given by r(t) = acos(wt) i + bsin(wt) j. Assume a and b are both positive and a > b. The plane polar coordinates of a particle at a time t equal to 1/8 of the time period T will be given by _
Homework Equations
r(t) = acos(wt) i + bsin(wt) j.
The...
the horizontal component would be (a-acosα) which is equal to 2asin2α/2 the vertical component 2asin(α/2)cos(α/2) the resultant would be 2asin(α/2)... I am sorry :P but this dosen't matches any of the options
ok I get what you are saying, so if we take A to be the acceleration of the block wrt to M. therefore it's acceleration with respect to the ground will be
(a - Acosα) horizontally and Asinα vertically. But now how do I find a relation between A and a
In the adjoining figure if acceleration of M with respect to ground is a, then
A) Acceleration of m with respect to M is a
B) Acceleration of m with respect to ground is asin(α/2)
C) Acceleration of m with respect to ground is a
D) Acceleration of m with respect to ground is atan(α)
The 2nd...