# Homework Help: Find hinge force

1. Mar 13, 2014

### utkarshakash

1. The problem statement, all variables and given/known data
A rod of mass m and length l is hinged about its end A and is vertical initially. Now the end A is accelerated horizontally with acceleration a=g m/s^2. The hinge reaction when the rod becomes horizontal is ?

2. Relevant equations
See the attached picture for relevant diagrams.

3. The attempt at a solution
When viewed from the frame of the hinge, the rod experiences a pseudo force = ma. Since the rod undergoes circular motion about the hinge A,

$R_x - ma = m \omega ^2 l$

where omega is the angular velocity of the rod when it becomes horizontal. Now I don't know what to write regarding the vertical forces. I'm sure they do not balance each other. Also how do I find omega at the instant?

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2. Mar 13, 2014

### paisiello2

Shouldn't l be l/2 in your equation?

I think the vertical forces do balance out so that should be easy.

My suggestion would be to use conservation of energy to get ω.

3. Mar 13, 2014

### utkarshakash

Why should it be l/2? If I balance the vertical forces I get R(y) = mg which is wrong according to the answer key.

4. Mar 13, 2014

### paisiello2

The center of mass is located at l/2.

You forgot to include the ma_y of the rod. If you take moments about the hinge then this will cancel with mg leaving R_y = 0.

5. Mar 13, 2014

### utkarshakash

But the rod undergoes circular motion about the hinge and not about its COM. R_y is not equal to zero, though.

6. Mar 13, 2014

### hav0c

Can you please clarify whether the rod is in equilibrium in its horizontal position or is it undergoing a circular motion.

7. Mar 14, 2014

### utkarshakash

The question says that the hinge is being moved. So, I think the rod should undergo circular motion about the hinge. The motion of the rod is a combination of rotational and translational motion when seen from ground. Am I wrong?

8. Mar 14, 2014

### Staff: Mentor

No. You're correct.

The first thing I always do on a problem like this is to focus on the kinematics of the motion. Let x(t) be the horizontal displacement of the hinge at time t, and let θ(t) be the counterclockwise angle that the rod makes with the vertical y direction. (At time zero, x(0) = 0 and θ(0) = 0.) In terms of x and θ, what are the coordinates xc and yc of the center of mass of the rod at time t? What are the x and y components of the velocity of the center of mass of the rod at time t? What are the x and y components of the acceleration of the center of mass of the rod at time t? In terms of θ, what is the angular velocity ω and the angular acceleration α of the rod?

Now you are ready to do force and moment balances on the rod. For an arbitrary value of θ, what are the force balances in the x and y directions? What is the moment balance around the center of mass of the rod?

Incidentally, happy $π$ day everyone.

Chet

9. Mar 14, 2014

### paisiello2

Yes, you are right, I'm sorry about that. What I should have said was:

ma_y = mlα/2

where α is the angular acceleration

Take moments about the hinge to solve for α and substitute into the equation for summing forces in the Y direction.

Last edited: Mar 14, 2014
10. Mar 14, 2014

### Staff: Mentor

The angular velocity around the center of mass is the same as the angular velocity around the hinge. Just consider the angle that the rod makes with the vertical direction at any point along the rod. It's the same at all locations along the rod.

Chet

11. Mar 14, 2014

### paisiello2

Yes, but taking moments about the hinge eliminates the reaction forces and makes the math simpler I think.

12. Mar 14, 2014

### Staff: Mentor

OK. Let's see how it plays out. But, don't forget that, for a rigid body, the F = ma equation only applies to the acceleration of the center of mass.

Chet

13. Mar 14, 2014

### utkarshakash

I could easily get R_y by your method but I'm still having trouble figuring out R_x. Here's what I did:

Initial Potential Energy of the rod = mgl/2 (assuming hinge to be at zero potential)
Let's say the rod acquires a linear velocity v when it becomes horizontal and angular velocity ω.

Final Energy = $\dfrac{ml^2 \omega ^2}{6} + \dfrac{mv^2}{2}$

But the problem is that there is no simple relation between v and ω and thus, I'm left with two variables.

14. Mar 14, 2014

### paisiello2

I think we can set v_y = 0 considering pure rotation.

Last edited: Mar 14, 2014
15. Mar 14, 2014

### Staff: Mentor

The moment balance on the rod gives:
$$mg\frac{l}{2}sinθ=Iα=\left(\frac{ml^3}{3}\right)\frac{d^2θ}{dt^2}$$
where θ is the angle of the rod measured clockwise from the vertical. Rearranging this gives:
$$\frac{d^2θ}{dt^2}=\frac{3g}{2l}\sinθ$$
If we multiply both sides of this equation by dθ/dt, and integrate with respect to t, we obtain:
$$\left(\frac{dθ}{dt}\right)^2=\frac{3g}{l}(1-\cosθ)$$
When θ=π/2, these equations reduce to:

$$α=\frac{3g}{2l}$$
$$ω^2=\frac{3g}{l}$$

I think these results might help.

Chet

16. Mar 14, 2014

### paisiello2

Not necessary to do all the integration and stuff. Just take moments directly when θ=90° and you get the same answer.

17. Mar 14, 2014

### Staff: Mentor

Really. That's interesting. How did you get the equation for ω2 without integrating?

Chet

18. Mar 14, 2014

### paisiello2

I see what you mean. I guess you are solving a differential equation then.

19. Mar 14, 2014

### Staff: Mentor

Yes, exactly.

This could also be viewed as a balance between rotational kinetic energy of the rod and gravitational potential energy of the rod, although, to be perfectly frank, it isn't obvious to me how this can be separated in some rational way from the overall mechanical energy balance on the rod. That's an issue that you and the OP were grappling with.

Chet

20. Mar 14, 2014

### utkarshakash

Shouldn't the torque due to pseudo force = ma be included as well?