NCERT Solutions

We need to find the magnitude and direction of acceleration for a stone undergoing uniform horizontal circular motion.

Concept: Centripetal acceleration

The centripetal acceleration is the acceleration of an object moving in a circular path, directed toward the center. Its magnitude is given by:

$$a_c = \omega^2 r$$

where $\omega$ is the angular velocity and $r$ is the radius. The angular velocity is related to the frequency $f$ by $\omega = 2\pi f$.

Given:

• Radius of the circle, $r = 80\,\text{cm} = 0.8\,\text{m}$
• Number of revolutions, $n = 14$
• Time interval, $t = 25\,\text{s}$

Solving:

1. Calculate the frequency of revolution, $f$

$$\begin{aligned} f &= \frac{n}{t} \\ &= \frac{14}{25}\,\text{s}^{-1} \\ &= 0.56\,\text{Hz} \end{aligned}$$

2. Calculate the angular velocity, $\omega$

$$\begin{aligned} \omega &= 2\pi f \\ &= 2 \cdot \pi \cdot 0.56 \\ &= 1.12\pi\,\text{rad}\cdot\text{s}^{-1} \end{aligned}$$

3. Calculate the magnitude of the centripetal acceleration, $a_c$

$$\begin{aligned} a_c &= \omega^2 r \\ &= (1.12\pi)^2 \cdot 0.8 \\ &= 1.2544\pi^2 \cdot 0.8 \\ &= 1.00352\pi^2 \end{aligned}$$

Using $\pi \approx 3.14159$:

$$\begin{aligned} a_c &\approx 1.00352 \cdot (3.14159)^2 \\ &\approx 9.90\,\text{m}\cdot\text{s}^{-2} \end{aligned}$$

4. Determine the direction

In uniform circular motion, the acceleration vector always points toward the center of the circular path.

Answer

$$9.90\,\text{m}\cdot\text{s}^{-2}$$

Direction: Towards the center of the circle.