Falling rod which is fixed at the pin

[issacnewton]

Hi

I am trying to solve this problem. I already got the answer but have some questions.

A long uniform rod of length L and mass M is pivoted
about a horizontal, frictionless pin through one end. The
rod is released from rest in a vertical position, as shown in
Figure P10.61. At the instant the rod is horizontal, find
(a) its angular speed, (b) the magnitude of its angular ac-
celeration, (c) the x and y components of the acceleration
of its center of mass, and (d) the components of the reac-
tion force at the pivot.

I got parts a,b,c and I am trying to do part d. Now I reasoned that, when the rod is rotating,
there will be two reactions at the pin. Since the motion of the rod at the pin is prevented in x and y directions, in free body diagrams, there will two forces, f₁
in an upward, y direction, and f₂ ,in the horizontal x direction.
And there will be weight, mg acting downwards at the center of mass. Let L be the length of the rod. Now from part a,b,c I got x and y components of the acceleration of the center of mass (the usual tangential acceleration).

$$a_{tx}=\frac{3g}{4}\sin \theta \cos \theta$$

where $$\theta$$ is the angle made by the rod with the vertical direction. so above acceeleration is directed to the right. And the downward y component is

$$a_{ty}=\frac{3g}{4} \sin^{2}\theta$$

I got the answer to the part d by setting up Newton's second law equations in x and y directions. But I tried to use another approach of setting up torque equations, and there was a problem. I reasoned that , angular acceleration, $$\alpha$$ would be same around any axis . The value of , $$\alpha$$ , I got was

$$\alpha=\frac{3g\sin \theta}{2L}$$

Now first let me write the values of f₁ and f₂ I got from solving
translational equations of motion for the center of mass.

$$f_1=mg\left[1-\frac{3}{4}\, \sin^2 \theta \right]$$

$$f_2=\frac{3mg}{4}\, \sin \theta \cos \theta$$

Now I set up the equation for rotational motion around the axis passing through the other free end of the rod.

$$\tau=(mg\sin\theta)\frac{L}{2}-(f_1 \sin \theta)L+(f_2 \cos \theta)L\cdots (1)$$

but $$\tau=I(-\alpha)$$

where I is the moment of inertia around the axis passing through the other end. So

$$\tau=-\frac{1}{3}mL^2\alpha$$

Now when I equate the two expressions, I get

$$\frac{1}{3}mL^2\alpha=(f_1 \sin \theta)L-(mg\sin\theta)\frac{L}{2}-(f_2 \cos \theta)L$$

Now I used the original value of alpha and simplified it.

$$\frac{3f_1}{mL}\sin \theta - \frac{3f_2}{mL}\cos \theta=\frac{3g\sin \theta}{L}$$

Now if I plug in the values of f₁ and f₂ in the above expression
, I end up with 1=0, which is wrong.

When I use the axes passing through the pin and the one passing through the center of mass
, I don't get this problem. When I choose an axis passing through any point beyond the
center of mass towards the other end, I get some problem. So I am suspecting that
there is something wrong in the equation 1. Can you see that ?

 

[tiny-tim]

hi IssacNewton!

… Now from part a,b,c I got x and y components of the acceleration of the center of mass (the usual tangential acceleration).

$$a_{tx}=\frac{3g}{4}\sin \theta \cos \theta$$

where $$\theta$$ is the angle made by the rod with the vertical direction. so above acceeleration is directed to the right. And the downward y component is …

i haven't looked through the rest of it, but you seem to have left out the https://www.physicsforums.com/library.php?do=view_item&itemid=27" of the centre of mass

 

[issacnewton]

Oh ya, I forgot that. With centripetal acceleration into the consideration, we get f₁
and f₂ as

$$f_1=mg-m\left[\frac{9g}{4}\sin^2\theta+\frac{3g}{2}(\cos\theta-1)\right]$$

and

$$f_2=\frac{3mg\sin\theta}{4}(3\cos \theta-2)$$

but my problem about that eq. 1 still holds

 

[tiny-tim]

When I use the axes passing through the pin and the one passing through the center of mass, I don't get this problem. When I choose an axis passing through any point beyond the center of mass towards the other end, I get some problem.

i think you're reversing the sign of α …

the angular velocity and acceleration are the same, no matter where they're measured about

 

[issacnewton]

I am just following signs properly. Since the rod's angular speed is increasing, and since its rotating in clockwise direction, the sign of angular acceleration should be negative ,so that $$\alpha$$ is just magnitude of the angular acceleration...

 

[tiny-tim]

I am just following signs properly. Since the rod's angular speed is increasing, and since its rotating in clockwise direction, the sign of angular acceleration should be negative ,so that $$\alpha$$ is just magnitude of the angular acceleration...

ah, you're measuring θ and α from the horizontal … i thought you were doing it from the vertical, because of the torque mgLsinθ …

have you mixed them up?

 

[issacnewton]

I will look into it.

Going to bed now :zzz: late night here

may be I can work whole thing in dream tonight