Angular momentum of the particle about point P as a function of time

[hhjjy]

Homework Statement

A particle of mass m moves in a circle of radius R at a constant speed v, as shown in Figure P11.4. If the motion begins at point Q at time t = 0, determine the angular momentum of the particle about point P as a function of time.

Relevant Equations

$$ L = \vec{r} \times \vec{p} $$

I don't understand why my solution is wrong.
Here is my solution.

$$ r_{\theta} = R\cos{\theta} \vec{i} + R\sin{\theta} \vec{j} $$
$$ v_{\theta} = v\cos(\theta + \frac{\pi}{2}) \vec{i} + v\sin(\theta + \frac{\pi}{2}) \vec{j} $$
$$ p_{\theta} = mvR(-\sin{\theta}) \vec{i} +mvR(\cos{\theta} \vec{j}) $$
$$ L_{\theta} = \vec{r} \times \vec{p}=mvR (-{\sin{\theta}}^2 + {\cos{\theta}}^2 )\vec{k} $$
$$ L_{\theta} = mvR(\cos{2\theta}) $$

Can someone explain for me?

 

[BvU]

And where is this point $P$ ?

 

[PeroK]

Where does time figure in this?

 

[jbriggs444]

It looks to me as if you are computing the cross product $\vec{r} \times \vec{p}$ as $r_x \cdot v_y + r_y \cdot v_x$. Should there be a minus sign in there? That would set things up for a different trig identity.

$$ p_{\theta} = mvR(-\sin{\theta}) \vec{i} +mvR(\cos{\theta} \vec{j}) $$
$$ L_{\theta} = \vec{r} \times \vec{p}=mvR (-{\sin{\theta}}^2 + {\cos{\theta}}^2 )\vec{k} $$

It would be good to write $\sin \theta^2$ as $\sin^2 \theta$ to avoid implying that $\theta$ is squared.

Then too, a moment's thought would discard the algebraic approach in favor of a simple geometric insight.

 

[vela]

Homework Statement:: A particle of mass m moves in a circle of radius R at a constant speed v, as shown in Figure P11.4. If the motion begins at point Q at time t = 0, determine the angular momentum of the particle about point P as a function of time.
Relevant Equations:: $$ L = \vec{r} \times \vec{p} $$

There should be an arrow over the $L$.

I don't understand why my solution is wrong.
Here is my solution.

$$ r_{\theta} = R\cos{\theta} \vec{i} + R\sin{\theta} \vec{j} $$
$$ v_{\theta} = v\cos(\theta + \frac{\pi}{2}) \vec{i} + v\sin(\theta + \frac{\pi}{2}) \vec{j} $$

There should be arrows over $r$ and $v$ on the lefthand sides.

$$ p_{\theta} = mvR(-\sin{\theta}) \vec{i} +mvR(\cos{\theta} \vec{j}) $$

Where did the factor of $R$ come from?

$$ L_{\theta} = \vec{r} \times \vec{p}=mvR (-{\sin{\theta}}^2 + {\cos{\theta}}^2 )\vec{k} $$
$$ L_{\theta} = mvR(\cos{2\theta}) $$

What happened to the $\vec k$?

 

[Charles Link]

There is a missing minus sign on the $-\sin^2{\theta}$ in the cross product, so that it will be $+\sin^2{\theta}$.

 

[hhjjy]

the figure is this one.

Where does time figure in this?

 

[TSny]

The diagram really helps. It shows that point $P$ is not at the center of the circle. The vector $\vec r$ in the equation $\vec L = \vec r \times \vec p$ is the position of the the particle relative to point $P$. So, $\vec r \neq R\cos(\theta) \hat i + R\sin(\theta) \hat j$

 

[hhjjy]

I

It looks to me as if you are computing the cross product $\vec{r} \times \vec{p}$ as $r_x \cdot v_y + r_y \cdot v_x$. Should there be a minus sign in there? That would set things up for a different trig identity.It would be good to write $\sin \theta^2$ as $\sin^2 \theta$ to avoid implying that $\theta$ is squared.

Then too, a moment's thought would discard the algebraic approach in favor of a simple geometric insight

How to use geometric sight to solve this problem? I have no idea when I first thought, so I use the algebraic approach to solve, but the result looks weird.

There is a missing minus sign on the $-\sin^2{\theta}$ in the cross product, so that it will be $+\sin^2{\theta}$.

According to Charles, I made some mistakes. If I correct the calculation mistake, It looks weird.

$\vec{L}(\theta) = mvR(\sin^2(\theta) + \cos^2(\theta)) --(1)$

We can use this function $\sin^2{\theta} + \cos^2 {\theta} = 1 --(2)$

to substitute.

The result is $L(\theta) = mvR$ which is totally irrelevant to the $\theta$

My thought is that if the result is related to $\theta$ I can import $\theta = \omega* times$.

 

[haruspex]

According to Charles, I made some mistakes.

You still seem to be making the mistake @TSny pointed out in post #8. What is the displacement vector from P to the mass as a function of time?

 

[jbriggs444]

How to use geometric sight to solve this problem?

I'd stay with the original description -- a particle moving in a circle at constant speed. About the origin, it has constant angular momentum. How can we determine its angular momentum about a parallel axis somewhere else?

 

[BvU]

Knowing where $P$ is located is really such a relief ! Now everything makes some sense!
And boy, are you getting some heavyweight help with this one

$\$

 

[hhjjy]

Oh , I got it . I miss a important idea.The point P is at (-R,0), so $\vec{r} = (R + R\cos{\theta}) \vec{i} + R \sin{\theta} \vec{j}$ . That makes sense if we take it back to the equation.
$\vec{L} = mvR(\sin^2{\theta}+\cos^2{\theta} + \cos{\theta}) = mvR(1 + \cos{\theta})\vec{k}$
By using this $\theta = \omega t = vRt$,the result is $\vec{L} = mvR(1+\cos{\frac{Vt}{R}})\vec{k}$.

 

[hhjjy]

I appreciate your help. Thank you everyone .

Knowing where $P$ is located is really such a relief ! Now everything makes some sense!
And boy, are you getting some heavyweight help with this one

##\ #

I'd stay with the original description -- a particle moving in a circle at constant speed. About the origin, it has constant angular momentum. How can we determine its angular momentum about a parallel axis somewhere else?