Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Problem:

A particle is constrained to move in a circle with a 10-meter radius. At one instant, the particle's speed is 10 meters per second and is increasing at a rate of 10 meters per second squared.

The book's answer:

The angle between the particle's velocity and acceleration vectors is given $45^0$.

My answer:

But I got $90^0$ by using this formula, $$v= r \omega \sin\theta$$ where $\omega= \sqrt{\frac{a}{r}}$

Where am I wrong?

share|improve this question
    
I don't understand your formula , can you give it's origin? –  Mr.ØØ7 May 10 '13 at 15:27
    
Search for centripetal acceleration –  Sonia Afrin May 10 '13 at 15:31
    
This is the second simple question regarding your homework… You should consider reading your book and lecture notes, since you won't learn that much if you let your problems solved by others. –  septi May 10 '13 at 16:10

2 Answers 2

up vote 3 down vote accepted

enter image description here

There are two perpendicular components of acceleration.

1) $a_t$ along the direction of velocity,that increase the speed. so, $a_t=10 m/s^2$

2)$a_c$ centripetal acceleration,towards the center of rotation . $a_c=\dfrac{v^2}r=10 m/s^2$

So, net $\vec a=\vec a_t +\vec a_c$

$|a_c|=|a_t|$, so it's equally inclined(at $45^0$) to both components.

Now you get your answer.

share|improve this answer

In this case, the particle is also having an angular acceleration, that is, $\vec{\omega}$ is not a constant. Since $\vec{v} = \vec{\omega} \times \vec{r}$, $\vec{a} = \vec{\alpha} \times \vec{r} + \vec{\omega} \times \vec{v}$, where $\vec{\alpha} = \dot{\vec{\omega}}$ is the angular acceleration. Therefore, $\vec{a}\cdot\vec{v} = \vec{\alpha}\times\vec{r}\cdot\vec{v}$. Angular acceleration is parallel to angular velocity, therefore, $\vec{\alpha} \times \vec{r} = r\alpha \hat{e}_{\theta}$, $\hat{e}_{\theta}$ being a unit vector in the direction of increasing angle. Since $\vec{v} = v\hat{e}_{\theta}$, $\vec{\alpha}\times\vec{r}\cdot\vec{v} = r\alpha v$.

We thus have $av\cos\theta = \alpha r v$. We know $|\vec{\alpha} \times \vec{r}| = 10 m/s^2$, $v = 10 m/s$ and $r = 10m$. For this combination of values,$ |\vec{a}| = 10\sqrt{2}$ and $|\vec{\alpha}| = 1$. Therefore, the equation $av\cos\theta = \alpha r v$ gives $\theta = \pi/4$.

In order to derive these relations, you may find it useful to have a cylindrical coordinate system with unit vectors $\hat{r}$, $\hat{\theta}$ and $\hat{z}$. Their orthogonality relations are $\hat{r}\times\hat{\theta} = \hat{z}$, $\hat{\theta}\times\hat{z} = \hat{r}$ and $\hat{z}\times\hat{r} = \hat{\theta}$. $\vec{v} = v\hat{\theta}$ while $\vec{\omega}$ and $\vec{\alpha}$ are along $\hat{z}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.