A particle moves in a plane with constant radial velocity $\dot{r} = 4 \ \frac{m}{s}$. The angular velocity is constant and has magnitude $\dot{\theta} = 2 \ \frac{rad}{s}$. When the particle is $3 \ m$ from the origin, find the magnitude of (a) the velocity and (b) the acceleration.

So $\bold{v} = v_{r} \bold{\hat{r}} + v_{\theta} \bold{\hat{\theta}}$ where the first component is the radial velocity and the second component is the angular velocity.

So we want units in $\frac{m}{s}$ (same units). So would I just change the $2 \ \frac{rad}{s}$ into $\frac{m}{s}$ and take the magnitude?

If we have constant velocity (both radial and angular) wouldn't the magnitude of the acceleration be $0$?

2. I think $||\bold{v}|| = \sqrt{36} \ \frac{\text{m}}{\text{s}}$ ?

3. is this too difficult?

4. Originally Posted by shilz222
A particle moves in a plane with constant radial velocity $\dot{r} = 4 \ \frac{m}{s}$. The angular velocity is constant and has magnitude $\dot{\theta} = 2 \ \frac{rad}{s}$. When the particle is $3 \ m$ from the origin, find the magnitude of (a) the velocity and (b) the acceleration.

So $\bold{v} = v_{r} \bold{\hat{r}} + v_{\theta} \bold{\hat{\theta}}$ where the first component is the radial velocity and the second component is the angular velocity.

So we want units in $\frac{m}{s}$ (same units). So would I just change the $2 \ \frac{rad}{s}$ into $\frac{m}{s}$ and take the magnitude?

If we have constant velocity (both radial and angular) wouldn't the magnitude of the acceleration be $0$?
The velocity is the vector sum of the radial and angular components. That is to say
$\vec{v} = \dot{r} \bold{\hat{r}} + r \dot{\theta} \bold{\hat{\theta}}$

Take the time derivative of this and you get:
$\vec{a} =(\ddot{r} - r \dot{\theta}^2) \bold{\hat{r}} + (2\dot{r} \dot {\theta} + r \ddot{\theta})\bold{\hat{\theta}}$

Note the "r" next to the $\theta$ in the velocity equation. This is where your units were inconsistent. The second term in the expansion for the acceleration is the centripetal acceleration term, which will be present even if there is no radial or angular accelerations. (The "complicated" form of the acceleration is due to the change in the unit vector $\hat{\theta}$ over time.)

-Dan