Angle between acceleration and velocity

Question

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,  $$v= r omega sintheta$$
  where $omega= sqrt{frac{a}{r}}$

Where am I wrong?

ABC · Accepted Answer

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.

Amey Joshi · 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, $vec{a}cdotvec{v} = vec{alpha}timesvec{r}cdotvec{v}$. Angular acceleration is parallel to angular velocity, therefore, $vec{alpha} times vec{r} = ralpha hat{e}_{theta}$, $hat{e}_{theta}$ being a unit vector in the direction of increasing angle. Since $vec{v} = vhat{e}_{theta}$, $vec{alpha}timesvec{r}cdotvec{v} = ralpha v$.

We thus have $avcostheta = alpha r v$. 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}| = 10sqrt{2}$ and $|vec{alpha}| = 1$. Therefore, the equation $avcostheta = alpha r v$ 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 $hat{r}timeshat{theta} = hat{z}$, $hat{theta}timeshat{z} = hat{r}$ and $hat{z}timeshat{r} = hat{theta}$. $vec{v} = vhat{theta}$ while $vec{omega}$ and $vec{alpha}$ are along $hat{z}$.

Angle between acceleration and velocity

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