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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?

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I don't understand your formula , can you give it's origin? – ABC 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
up vote 3 down vote accepted

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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.

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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 $\hat{z}$.

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