Hi again,
Concerning the derivative of the work of the flywheel,
Work= ½ * I * ( W2^2 – W1^2)
d/dt work= ½ * I * d/dt (W2^2) –d/dt (W1^2)...(1)
d/dt (W^2)= d/dt( W.W)= d/dt (W) * W + W * d/dt(W) = acc*W + W*acc = 2 * acc * W
Back to (1)
d/dt (work)=1/2 * I * [ (2 * acc2 * W2)- (2 * acc1 * W1)]...
Thank you for your reply and for being patient with me.
While searching the internet for how to derive the work, I found a similar problem in a book, shown in the link below: (Example 10.11)...
Thanks for your reply..
Concerning the first part, to account for the flywheel's effect during engagement, is it correct if I calculate the power of the flywheel alone as dwork/dt and add it to the cycling power equation as:
Cycling power = Torque * Crank Angular Velocity (rad/s) + dwork/dt...
Dear Friends
I’m in need to solve a problem in physics, which might be easy for you but it is not for me as my background in physics is not good.
I have a bike (drawn in simulation platform) with a flywheel connected to the crank (or shaft). The flywheel is not always fixed (or engaged)...