Dynamics...Rigid body angular acceleration

[brk51]

Homework Statement

The 25-lb slender rod has a length of 6 ft. Using a collar of negligible mass, its end A is confined to move along the smooth circular bar of radius 32√ ft. End B rests on the floor, for which the coefficient of kinetic friction is μB = 0.24. The bar is released from rest when θ = 30∘.

Homework Equations

lots of em

The Attempt at a Solution

I wanted to do this in normal-tangential coordinates b/c it seemed easier but I'm getting the wrong answer.. here is my work .

Let me know if any of my logic here is wrong.
---The bar is released from rest so the sum of the forces should equal 0 but NOT the moments.
----The sum of forces in the tangent direction also = 0 because their is no acceleration in that direction b/c the object is constrained to moving in a circle
---I'm summing the moments about point G, the center of mass for the rod, thus the the equation becomes
Sum of (M_G) = (I_G)* Alpha
--- Normal Force at A crosses point G so no moment there. Same goes for weight.
--- thus only the frictional force and the normal force's tangent components are present in the moment equation..

Please help a guy out! Thanks in advance.

 

[NFuller]

The bar is released from rest so the sum of the forces should equal 0 but NOT the moments.

The bar will accelerate as it begins to fall so the sum of the forces cannot be zero.

The sum of forces in the tangent direction also = 0 because their is no acceleration in that direction

There is acceleration in that direction.

Try and work with the equation relating torque and angular acceleration with the pivot point at the END of the rod, not its center of mass.

 

[brk51]

So when is there never tangential acceleration hypothetically.

And if i was to take the moments about the end of the rod. It would have to equal the kinetic moments of the rod..(Ma_t + Ma_n) ?

 

[brk51]

So when is there never tangential acceleration hypothetically.

And if i was to take the moments about the end of the rod. It would have to equal the kinetic moments of the rod..(Ma_t + Ma_n) ?

+ I_end * alpha

 

[NFuller]

The first thing you should do is figure out what the x motion of the end of the rod is against the floor as a function of $\theta$. This can be a little tricky, but it can be done with right triangle trigonometry. Next, work out the force of friction on the end of the rod as a function of $\alpha$, this force must also appear at the junction of the rod and rail since the rail is what is pushing the rod against the floor. This force can be cast as a torque applied at the end of the rod apposing the torque applied by gravity. Once you have these torques, it is a simple matter to find the angular acceleration.

And if i was to take the moments about the end of the rod. It would have to equal the kinetic moments of the rod..(Ma_t + Ma_n) ?

No, this would involve the parallel axis theorem https://en.wikipedia.org/wiki/Parallel_axis_theorem

 

[brk51]

The first thing you should do is figure out what the x motion of the end of the rod is against the floor as a function of $\theta$. This can be a little tricky, but it can be done with right triangle trigonometry. Next, work out the force of friction on the end of the rod as a function of $\alpha$, this force must also appear at the junction of the rod and rail since the rail is what is pushing the rod against the floor. This force can be cast as a torque applied at the end of the rod apposing the torque applied by gravity. Once you have these torques, it is a simple matter to find the angular acceleration.No, this would involve the parallel axis theorem https://en.wikipedia.org/wiki/Parallel_axis_theorem

Im so confused...

 

[devTr4P]

Im so confused...View attachment 206633

In the first image I have drawn the force-couple equivalency. Basically, the sum of forces equals the mass * acceleration of center of mass vector and the sum of torques equals the moment of inertia with respect to G * angular acceleration couple PLUS the torque produced by the mass * acceleration vector.

In the second image I have drawn the kinematics diagram. It basically states the acceleration of the center of mass (aX, aY). This acceleration is equal to the acceleration of B plus the acceleration of G relative to B. Similarly, the acceleration of B is equal to the acceleration of A plus the acceleration of B relative to A.
The acceleration of A can be found as a function of phi dot dot (angular acceleration of phi) and phi dot dot can be found as a function of alpha (angular acceleration of the rod) by using trig.