Question

A particle travels in a circular orbit of radius 21 m. Its speed is changing at a rate of 23.1 m/s2 at an instant when its speed is 37.2 m/s. What is the magnitude of the acceleration (in m/s?) of the particle?

Answer 1

The particle has an acceleration vector with one component directed toward the center of its orbit, and the other directing tangentially to its orbit. Call these components $\vec a_c$ ($c$ for center) and $\vec a_t$ ($t$ for tangent). Then its acceleration vector has magnitude

$|\vec a|=\sqrt{\|\vec a_c\|^2+\|\vec a_t\|^2}$

We have

$\|\vec a_c\|=\dfrac{\|\vec v\|^2}r$

where $\|\vec v\|$ is the particle's speed and $r$ is the radius of orbit, so

$\|\vec a_c\|=\dfrac{\left(37.2\frac{\rm m}{\rm s}\right)^2}{21\,\rm m}=65.9\dfrac{\rm m}{\mathrm s^2}$

We're given that the particle's speed changes at a rate of 23.1 m/s^2. Its velocity vector points in the same direction as $\vec a_t$, i.e. perpendicular to $\vec a_c$, so

$\|\vec a_t\|=23.1\dfrac{\rm m}{\mathrm s^2}$

Then the magnitude of the particle's acceleration is

$\|\vec a\|=\sqrt{\left(65.9\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(23.1\dfrac{\rm m}{\mathrm s^2}\right)^2}=\boxed{69.8\dfrac{\rm m}{\mathrm s^2}}$