Question

A sprinter running on a circular track has a velocity of constant magnitude 9.20 m/sand a centripetal acceleration of magnitude$\mathbf{3}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. What are (a) the track radius and (b) the period of the circular motion?

Short answer

a) The track radius is 22.3 m.

b) The period of circular motion is 15.2 s.

Step-by-step solution

Given information

It is given that,

$$\begin{gathered} v=9.20\mathrm{m}/\mathrm{s} \\ {\mathrm{a}}_{\mathrm{c}}=3.80\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Determining the concept of centripetal acceleration

This problem deals with the centripetal acceleration. Centripetal acceleration is the acceleration of a body traversing a circular path. The sprinter is running on a circular track, so the net force will be the centripetal force. The radius and period of sprinter using the equation of centripetal acceleration and period in circular motion can be found.

Formulae:

The centripetal acceleration is given by,

$$\begin{gathered} a_c=\frac{v^2}{r} \\ v=\frac{2\pi r}{T} \end{gathered}$$ (i)

(ii)

Where, r is radius, v is velocity, Tis time and a is an acceleration.

(a) Determining the radius of circular track

Using equation (i),

$$\begin{gathered} r=\frac{v^2}{a_c} \\ =\frac{{\left(9.20\right)}^2}{\left(3.80\right)} \\ =22.27\mathrm{m}\mathrm{or}22.3\mathrm{m} \end{gathered}$$

(b) Determining the period of sprinter

Using equation (ii),

Time will be,

$$\begin{gathered} T=\frac{2\pi r}{v} \\ =\frac{\left(2\times 3.14\times 22.27\right)}{\left(9.20\right)} \\ =15.20\mathrm{s} \end{gathered}$$