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