Acceleration of point

The mass point moves evenly along a circle with a radius of 1.2 m at an angular velocity of 25 rad/s. Determine the frequency, period, and centripetal acceleration!

Correct answer:

f =  3.9789 Hz
T =  0.2513 s
a =  750 m/s²

Step-by-step explanation:

$$\begin{gathered} r=1.2\text{m} \\ \omega =25\text{rad/s} \\ \omega =2\pi f \\ f=\omega /(2\pi )=25/(2\cdot 3.1416)=3.9789\text{Hz} \end{gathered}$$

$T=\frac{1}{f}=\frac{1}{3.9789}=0.2513\text{s}$

$$\begin{gathered} v=\omega \cdot r=25\cdot 1.2=30\text{m/s} \\ a=\frac{v^2}{r}=\frac{30^2}{1.2}=750{\text{m/s}}^2 \end{gathered}$$

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