Question

A boy whirls a ball on a string along a horizontal circle of radius 98 cm. The angular velocity (in rad $s^{-1}$) so that its centripetal acceleration has the same magnitude as $g$ is ________.

• A. $\sqrt{10}$
• B. $\frac{1}{\sqrt{10}}$
• C. 10
• D. 0.1
• E. 100

Hint:

$\omega = \sqrt{a/r}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The acceleration of an object moving in a circle at a constant angular velocity is directed towards the center of the circle. This is called centripetal acceleration. The problem states that the magnitude of this acceleration is equal to the magnitude of the acceleration due to gravity, \(g\).
Step 2: Key Formula or Approach
The formula for centripetal acceleration (\(a_c\)) in terms of angular velocity (\(\omega\)) and radius (\(r\)) is:
\[ a_c = \omega^2 r \] We are given the condition \(a_c = g\). We need to solve for \(\omega\).
\[ \omega^2 r = g \implies \omega = \sqrt{\frac{g}{r}} \] Step 3: Detailed Explanation
1. List the given values in SI units.
- Radius, \(r = 98 \text{ cm} = 0.98 \text{ m}\).
- Acceleration due to gravity, \(g\). We can use the approximation \(g \approx 9.8 \text{ m/s}^2\).
2. Set up the equation.
We are given that the centripetal acceleration equals the acceleration due to gravity:
\[ a_c = g \] \[ \omega^2 r = g \] 3. Solve for the angular velocity \(\omega\).
\[ \omega^2 = \frac{g}{r} \] Substitute the values:
\[ \omega^2 = \frac{9.8}{0.98} \] \[ \omega^2 = \frac{980}{98} = 10 \] Now, take the square root to find \(\omega\):
\[ \omega = \sqrt{10} \, \text{rad s}^{-1} \] Step 4: Final Answer
The required angular velocity is \(\sqrt{10}\) rad s\(^{-1}\).