Question

The centrifugal force (N) acting on a material mass 2g present at the tip of a beater of radius 25 cm rotating at 600 rpm is

• A. \( 0.1 \pi^2 \)
• B. \( 0.2 \pi^2 \)
• C. \( 0.005 \pi^2 \)
• D. \( 0.05 \pi^2 \)

Hint:

Always convert units to the standard SI units (kg, meters, radians/second) before calculating centrifugal force. Remember the formula: \( F = m \omega^2 r \).

Answer 1

The Correct Option is B

The centrifugal force \(F\) is given by the formula:

\(F = m \omega^2 r \) 

where:

• \(m\) is the mass (in kg)
• \(\omega\) is the angular velocity (in rad/s)
• \(r\) is the radius (in meters)

First, convert the given values to the correct units:

• \(m = 2g = 0.002 kg\)
• \(r = 25 cm = 0.25 m\)
• \(\omega = (600 \, \text{rpm}) \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = 20\pi \, \text{rad/s}\)

Now, substitute these values into the formula:

\(F = (0.002 \, \text{kg}) \times (20\pi \, \text{rad/s})^2 \times (0.25 \, \text{m}) \)

\(F = 0.002 \times 400\pi^2 \times 0.25 \)

\(F = 0.2\pi^2 \, \text{N}\)