## Statement

A 1 m radius circular disk rotates at constant angular velocity, so that it takes 1.2 s for a full rotation. What is the normal or centripetal acceleration of the external points of its periphery?

## Solution

Since the angular velocity is constant and the trajectory of any point is circular we have a uniform circular motion or u.c.m.

Data

R = 1 m

T = 1.2 s

Resolution

If it takes 1.2 s for a full rotation, the disk turns with a period T=1.2 s. From this period, we can calculate the angular velocity of the disk:

$$\omega =\frac{2\pi}{T}\Rightarrow \phantom{\rule{0ex}{0ex}}\omega =\frac{2\pi}{1.2}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)\omega =1.67\pi rad/s}$$

Considering the definition of normal or centripetal acceleration, we can calculate its value using the following expression:

$${a}_{n}=\frac{{v}^{2}}{R}={\omega}^{2}\xb7R\Rightarrow \phantom{\rule{0ex}{0ex}}{a}_{n}=(1.67\pi {)}^{2}\xb71\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{){a}_{n}=27.52m/{s}^{2}}$$