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?
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.
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 $\hat{z}$.
