Let us analyze and solve the given problem.

The question involves determining the centripetal acceleration of an object traveling in a horizontal circular path at constant speed.

Given:

• Radius of the circular path: $r$
• Period of revolution: $T$

We know the formula for centripetal acceleration, $a_c$, which is given by:

$a_c = \frac{v^2}{r}$

Here, $v$ is the speed of the object. Since the object completes one revolution in time $T$, the speed $v$ can be expressed as:

$v = \frac{\text{circumference}}{\text{period}} = \frac{2\pi r}{T}$

Now substitute $v = \frac{2\pi r}{T}$ into $a_c = \frac{v^2}{r}$:

$a_c = \frac{\left(\frac{2\pi r}{T}\right)^2}{r}$

Simplify:

$a_c = \frac{4\pi^2 r^2}{T^2 \cdot r}$

$a_c = \frac{4\pi^2 r}{T^2}$

Answer:

The centripetal acceleration is:

$\boxed{\frac{4\pi^2 r}{T^2}}$

Thus, the correct choice is (C).

Do you have any further questions, or would you like me to clarify any steps?

Related Questions:

1. How do we calculate the centripetal force acting on the object in this situation?
2. If the period $T$ were halved, how would that affect the centripetal acceleration?
3. What is the relationship between centripetal acceleration and angular velocity?
4. Why is the centripetal acceleration independent of the mass $m$ of the object?
5. Can you describe a real-life example of centripetal acceleration?

Tip:

Always remember that centripetal acceleration depends only on the radius of the circular path and the square of the speed (or period of revolution). It points toward the center of the circle!