Rigid body dynamics (kinematics of a point)

[xzibition8612]

Homework Statement

See attachment "problem"

Homework Equations

The Attempt at a Solution

See attachment "work" and "answer"

I got 0.92, but the book gives 1.73. Something went wrong, but I can't find it.

 

[tiny-tim]

hi xzibition8612!

why have you made x_A negative?

 

[xzibition8612]

xA is negative because if you look at the attachment "question", I set point C as the origin. Left of C is in the negative direction. Thanks for responding.

 

[RTW69]

I think you have made some sign errors. Check your quadratic equation, I don't think q is correct.

 

[RTW69]

You made Xa positive the first time you used the law of cosines. Move the axis to A and see if you get a better result.

 

[RTW69]

My error, after deciding to do the math, your length of q is 13.9 m but I agree with tiny-tim, I think X_a= 10 m. The curious thing is that dK_b/dt is -1.73. Right magnitude, wrong direction

 

[xzibition8612]

goddamn this problem is driving me insane, i set origin at A and the answer turns out to be negative.

 

[RTW69]

I think the negative sign means that point B is decelerating. It is does not mean that V_b is pointing toward c.

There is another way to do this problem just using vectors and the kinematics of rigid bodies. The plane motion of the rigid bar can replaced by a translation defined by the motion of an arbitrary reference point and by a rotation about that point.

The vector equation is V_B= V_B/A + V_A

You know the magnitude and direction of V_A, the direction of V_B. V_B/A is the vector perpendicular to the rigid bar at Pt. B causing the rigid bar to rotate CCW. You can determine the angle of V_B/A from the horizontal by geometry. Through vector addition you can determine the magnitude of V_B. You will see that it's magnitude is 1.73 m/s away from B. So forget the negative sign, it is not a direction sign.

 

[xzibition8612]

alright man thanks a lot