How to Solve Hinge Force Problems?

[mmaismma]

Homework Statement

In the figure shown a uniform rod of mass m and length l is hinged. The rod is released when the rod makes angle θ with the vertical.

Relevant Equations

Find
A) The angular velocity of the rod at the lowest position.
B) Normal Reaction due to the hinge just after the rod is released

I have solved problem (A).

But I couldn't solve problem (B).
Here is my attempt:

 

[haruspex]

Homework Statement:: In the figure shown a uniform rod of mass m and length l is hinged. The rod is released when the rod makes angle θ with the vertical.
Homework Equations:: Find
A) The angular velocity of the rod at the lowest position.
B) Normal Reaction due to the hinge just after the rod is released

View attachment 254984

I have solved problem (A).
View attachment 254982

But I couldn't solve problem (B).
Here is my attempt:View attachment 254983

The equations you have written in part B seem to be statics equations, but the rod will be accelerating.

 

[mmaismma]

The equations you have written in part B seem to be statics equations, but the rod will be accelerating.

You are absolutely right. Here is my attempt. Please verify the solution.

 

[haruspex]

You have a few sign errors.
Which ways are you taking as positive for a_t?
And which way does centripetal acceleration act? (But that error didn’t matter because it is zero.)
You also seem to have dropped the mg sin theta term in Ny.

 

[mmaismma]

You have a few sign errors.
Which ways are you taking as positive for a_t?
And which way does centripetal acceleration act? (But that error didn’t matter because it is zero.)
You also seem to have dropped the mg sin theta term in Ny.

The signs are denoted in this image. And I haven't dropped the mgsin theta term. Also the a_t is taken in the +ve y-axis side (i.e. along N_y).

 

[haruspex]

The signs are denoted in this image. And I haven't dropped the mgsin theta term. Also the a_t is taken in the +ve y-axis side (i.e. along N_y).View attachment 255076

Ok, so a_t is positive up, but then which way are you taking α as positive, and what then is the correct relationship between them?

Wrt the mg sin(theta) term you have not underscored the line I had in mind. (This is an example of why the guidelines say not to post working as images. If you take the trouble to type it in it is much easier to refer to. Or you could number your equations.)
Look at the next two lines. Nothing missing?

 

[mmaismma]

Ok, so a_t is positive up, but then which way are you taking α as positive, and what then is the correct relationship between them?

Wrt the mg sin(theta) term you have not underscored the line I had in mind. (This is an example of why the guidelines say not to post working as images. If you take the trouble to type it in it is much easier to refer to. Or you could number your equations.)
Look at the next two lines. Nothing missing?

Sorry. My bad. Here is my revised attempt. Please check.
$\begin{align} \Rightarrow F_{net}{}_{x} &=ma_x\nonumber\\ \Rightarrow mg\sin \theta - N_y &= ma_t \nonumber\\ \Rightarrow N_y &= mg\sin \theta - ma_t \nonumber\\ &= mg\sin\theta -m \times \frac {3g\sin\theta} {2l} \times \frac {l} {2} \nonumber\\ \Rightarrow \mathbf {N_y = \frac {mg\sin \theta}{4}}\end{align}$
And obviously $\begin{align}\mathbf{N_x=mg\cos\theta}\end{align}$

 

[haruspex]

Sorry. My bad. Here is my revised attempt. Please check.
$\begin{align} \Rightarrow F_{net}{}_{x} &=ma_x\nonumber\\ \Rightarrow mg\sin \theta - N_y &= ma_t \nonumber\\ \Rightarrow N_y &= mg\sin \theta - ma_t \nonumber\\ &= mg\sin\theta -m \times \frac {3g\sin\theta} {2l} \times \frac {l} {2} \nonumber\\ \Rightarrow \mathbf {N_y = \frac {mg\sin \theta}{4}}\end{align}$
And obviously $\begin{align}\mathbf{N_x=mg\cos\theta}\end{align}$

Looks good.

 

[mmaismma]

Looks good.

It doesn't match my anwer key. Probably the answer key is wrong. But can you ensure me if my answer key is wrong. Here is what my answer key says:
$$N_1=\frac{3mg\sin\theta\cos\theta}{4}\\
N_2=mg\left[1-\frac{3\sin{}^2\theta}{4}\right]$$

 

[haruspex]

It doesn't match my anwer key. Probably the answer key is wrong. But can you ensure me if my answer key is wrong. Here is what my answer key says:
$$N_1=\frac{3mg\sin\theta\cos\theta}{4}\\
N_2=mg\left[1-\frac{3\sin{}^2\theta}{4}\right]$$

Same answer but using vertical and horizontal coordinates.

 

[mmaismma]

$$\begin{align}N_1&=N_y\cos\theta+N_x\cos\theta\nonumber\\
&=\frac{3mg\sin\theta\cos\theta}4+mg\cos{}^2\theta\neq\frac{3mg\sin\theta\cos\theta}4\nonumber\end{align}$$

 

[haruspex]

$$N_1=N_y\cos\theta+N_x\cos\theta$$

That doesn't strike you as unusual?

 

[mmaismma]

That doesn't strike you as unusual?

What did't strike me?
Since I have got to the answer, You can now post the complete answer as per Physics forums rules.

Giving Full Answers:
On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Complete solutions can be provided to a questioner after the questioner has arrived at a correct solution. If the questioner has not produced a correct solution, complete solutions are not permitted, whether or not an attempt has been made.

 

[haruspex]

What did't strike me?
Since I have got to the answer, You can now post the complete answer as per Physics forums rules.

It would be most odd that in converting between coordinate systems you would use the cosine of the angle in both terms of the one equation. If you do the right conversions you will find your answer and the book answer are the same.

 

[mmaismma]

You are right it ain't striking me. I can't understand what are you trying to say.

 

[haruspex]

You are right it ain't striking me. I can't understand what are you trying to say.

What is the vertical component of your N_x, taking up as positive?
What is the vertical component of your N_y, taking up as positive?
They cannot both involve cos(θ). One of them must be sine. And I believe one of them is negative.

 

[mmaismma]

You are right my bad. But it still doesn't match my answer key:
$\begin{align}&N_x\cos\theta+N_y\sin\theta=N_1\nonumber\\&mg\cos\theta\times\cos\theta+\frac{mg\sin\theta}4\sin\theta\nonumber\\&\frac{mg(4\cos{}^2\theta+\sin{}^2\theta)}4\neq N_1\nonumber\end{align}$

 

[haruspex]

You are right my bad. But it still doesn't match my answer key:
$\begin{align}&N_x\cos\theta+N_y\sin\theta=N_1\nonumber\\&mg\cos\theta\times\cos\theta+\frac{mg\sin\theta}4\sin\theta\nonumber\\&\frac{mg(4\cos{}^2\theta+\sin{}^2\theta)}4\neq N_1\nonumber\end{align}$

I believe N₁ is horizontal, so you changed the wrong trig function for that case.

 

[mmaismma]

I believe N₁ is horizontal, so you changed the wrong trig function for that case.

You are right. Continuing previous equation:
$\begin{align}&=mg\frac{1+3\cos^2\theta}4\nonumber\\&=mg\frac{1+3-3\sin^2\theta}4\nonumber\\&=mg\frac{4-4\sin^2\theta}4\nonumber\\&=\mathbf{mg(1-\frac{sin^2\theta}4)}\end{align}$
And
$\begin{align}N_1&=N_y\cos\theta-N_x\sin\theta\nonumber\\&=\frac{mg\sin\theta}4\cos\theta-mg\cos\theta\times\sin\theta\nonumber\\&=\mathbf{\frac{-3mg\sin\theta\cos\theta}4}\\&=\text{answer in opposite direction}\nonumber\end{align}\\$
Thanks.

 

[haruspex]

You are right. Continuing previous equation:
$\begin{align}&=mg\frac{1+3\cos^2\theta}4\nonumber\\&=mg\frac{1+3-3\sin^2\theta}4\nonumber\\&=mg\frac{4-4\sin^2\theta}4\nonumber\\&=\mathbf{mg(1-\frac{sin^2\theta}4)}\end{align}$
And
$\begin{align}N_1&=N_y\cos\theta-N_x\sin\theta\nonumber\\&=\frac{mg\sin\theta}4\cos\theta-mg\cos\theta\times\sin\theta\nonumber\\&=\mathbf{\frac{-3mg\sin\theta\cos\theta}4}\\&=\text{answer in opposite direction}\nonumber\end{align}\\$
Thanks.

A couple of typos in the N₂ path. Should have ended up with $1-\frac{3\sin^2}4$

 

[mmaismma]

Yeah you are right.
\begin{align}&=mg\frac{1+3\cos^2\theta}4\nonumber\\&=mg\frac{1+3-3\sin^2\theta}4\nonumber\\&=mg\frac{4-3\sin^2\theta}4\nonumber\\&=\mathbf{mg(1-\frac{3sin^2\theta}4)}\end{align}