Problem statementSolution video |

**DISCUSSION THREAD**

**Discussion**

Note that the plate rotates about point O. Therefore, O is the center of the circular paths of points A and B. From the animation above, we see that the velocities of A and B are tangent to their circular paths, as expected. The accelerations of A and B are NOT perpendicular to the paths of A and B since the speeds of A and B are increasing in time (and consequently, A and B each have positive tangential components of acceleration).

Initially, the acceleration for these two points is nearly aligned with velocity, since the speeds are small and therefore the centripetal components of acceleration are small. Near the end of the first revolution of the plate, the speeds have increased to the point where the centripetal components of acceleration dominate, and acceleration is nearly perpendicular to the path.

**Solution hints**

For Part a) of this problem, it is recommended that you use the rigid body kinematics equations using point O as the reference point, since the velocity and acceleration of O are zero. That is, you should use v_B = v_O + Ω x r_B/O and a_B = a_O + Ω_dot x r_B/O - Ω^2*r_B/O. Repeat the process for finding the velocity and acceleration of A.

For Part b) of this problem, it is recommended that you use the rigid body kinematics equations with point A first. This will give you the equations needed to find Ω_dot. Then, use the rigid body kinematics equations to find the acceleration of B.

Any questions?? Please ask/answer questions regarding this homework problem through the "Leave a Comment" link above.

The hardest part of this problem is part B. Solve for the acceleration about point A with respect to O. Then, since you know the acceleration at point A is only in the j direction, you can set the i component of your findings to zero. Thus, you can solve for the actual value of omega_dot. Plug this value into the equation for acceleration at point B and you will have your final answer.

Yeah I agree with this, as always watch your cross products, as with these equations there are a lot of negatives and it is easy to forget one. I would always recommend writing them out to make sure you don't forget any

Can we assume for part A that the velocity of B is in the negative x direction and that the velocity of A is in the positive y direction since they are at the max range of y and x respectively?

Michael: There are really no "assumptions" to be made here. You should use the rigid body kinematics equation to find the velocities of points A and B.

For point B, it will work out that, yes, that v_B will be in the negative x-direction.

For point A, you will find out that v_A will have BOTH x- and y-components - it is not just in the positive y-direction.

You can see the directions for v_A and v_B in the above animations.

Just use the fundamental equations, and have confidence in the math!

Using Vb = Va + w x Rb/a, I end up with two identical equations for i and j. My thought is that I did Rb/a wrong, but I’m not sure where. I found positions relative to O for B and A, then did Rb - Ra. Is there something noticeable here that I did incorrectly?

Missed the point in the discussion saying to use O as the reference point. Figured it out, thanks!

In part B when solving for omega_dot using the rigid body equation for point A with respect to O, would it make sense to treat omega_dot as alpha? Assuming this is correct, omega_dot is in the k direction before doing the cross product?

Yes, the alpha vector will be Omega_dot*k_hat.

In A, I'm confused as to how a cross product of omega (units rad/s) X r (units m) gets resulting units of m/s. Shouldn't it mathematically be rad-m / s or am I missing something?

Radians are dimensionless. Therefore, rad*m/s is the same as m/s.

For part (b), i used the 2-D equation for acceleration at A but i am unsure where i can find omega_dot. Should i set a_A = 0 then solve or how can i get omega dot by itself?

You need to set a_A = -a_A*j_hat, NOT zero. Through the x- and y-components of your acceleration equation, you will have two equations in terms of two unknowns: a_A and alpha (-Omega_dot).

For part A, is omega_dot equal to the 10 rad/s^2 provided in the part-a find section or do we have to solve for omega_dot in some other way.

Where did the equation a_B = a_O + Ω_dot x r_B/O - Ω^2*r_B/O, come from? And can we not just use the equation a_B = a_O +Ω x{Ω x r_B/O] + Ω_dot x r_B/O. We should get the same answers right?

I found the equation I was asking about above.

Elijah: Please look at page 89 of the lecture book. See the fourth bullet point on that page to see the origins of the first equation that you wrote down. As you can see there, the two equations are the same, under the right conditions.