Acceleration and Tangential Velocity Vector?

Question

Find the resultant acceleration of a particle moving on a circle of radius $0.70\ m$, if its angular speed is 37 rpm and its tangential acceleration is $2.9 \frac m{s^2}$.

Express the angle with respect to the tangential velocity vector.

I thought that I need to start out finding the circumference fist by $2\pi \cdot 0.7 = 1.4\cdot \pi$.

Then take $\frac {1.4\cdot\pi\cdot37}{ 60\,sec} = 2.7122$.

However, this is wrong and I have no idea how to figure the tangential velocity vector. Can anyone help me out on this?

Answer 1

First, you need to find the centripetal acceleration of the particle. This is a function of the radius in meters and the speed in meters per second, which you've already computed. Do you know the formula that expresses centripetal acceleration in these terms?

Then, the resultant acceleration is a vector sum of the centripetal acceleration and the tangential acceleration. Can you draw the relevant diagram?

(Edit: fixed the units of speed, since you have meters per second.)

Answer 2

The acceleration has two perpendicular components. The tangential acceleration is given as $a_t=2.9m/s^2$. The centripetal acceleration (that keeps a body moving in a circle) is given by $a_c=\omega^2r$. Here $\omega$ is the angular velocity in radians per second. Every rotation has an angle of $2\pi$, so $\omega=2\pi\cdot 37/60\ rad/s$. The angle between the acceleration and the tangential vector is $\theta$. If you draw the two components you will see that $$\tan\theta=\frac{a_c}{a_t}$$ Calculate first $\omega$, then $a_c$, and finally plug them into the above equation.