Velocity,A​cceleratio​n,Angle

How do I find plots for

velocity of link3 vs angle of link1 Acceleration of link3 vs angle of link1.

[James Tursa]

Do you know the derivatives of "angle of Link1"? I.e., ph1dot and phi1dotdot? You will need these to determine the velocity and acceleration of Link3. Just take the derivative of your Link3 equation and then plug in those angle derivatives.

[James Tursa]

L3 = L1 * cos(phi1) + sqrt(L2^2 - (L1*sin(phi1))^2)

Then just take derivatives of this equation and plug in the derivatives of phi1 to get your results.

[James Tursa]

I can't open the image. Is this a new question?

[James Tursa]

And I don't know how to view the image.

[James Tursa]

I see the image below, but that doesn't change my comment above. You simply need to take the derivative of the L3 equation I gave you above and then plug in the phi1 derivatives to get your answer. Do you know how to take the derivatives?

Answer 1

[James Tursa]

This is a dynamics class. Dynamics involves derivatives. How can you be in a dynamics class without knowing calculus and how to take derivatives? All you have to do is know how to take the derivative of sqrt and trig functions and apply some chain rule stuff to get your answer. E.g., if you had the following for the x-coordinate of point B with respect to point A:

Bx = L1 * cos(phi1)

And you were asked to find the velocity of the x-coordinate of point B, then the derivative of Bx with respect to time would be

d(Bx)/dt = -L1 * sin(phi1) * d(phi1)/dt

And since they told you that phi1 is changing with constant angular rate w1, you get

d(Bx)/dt = -L1 * sin(phi1) * w1 <-- Velocity of Bx

If you were asked about the acceleration of Bx, then take another derivative:

d2(Bx)/dt^2 = -L1 * cos(phi1) * d(phi1)/dt * w1

And again knowing that d(phi1)/dt is the constant w1 you get

d2(Bx)/dt^2 = -L1 * cos(phi1) * w1^2 <-- Acceleration of Bx

This is simple calculus. You have to know this stuff to get by in a dynamics class. Just apply this method to the L3 equation I gave you above.

[Sachith Wijayanayaka]

Thank you sir