A man swung an object of mass 2 kg in a circular path with a rope 1.2 m long. If the object was swung at 120 rev/min, find the tension in the rope.

1. Analyze the question

We are given:

• Mass ($m$) = 2 kg
• Radius ($r$) = 1.2 m
• Frequency ($f$) = 120 revolutions per minute (rev/min)
• We need to find the tension ($T$) in the rope, which acts as the centripetal force ($F_c$).

2. Explain the concept

The tension in a rope swinging an object in a horizontal circle is equal to the centripetal force. The formula for centripetal force is:

Where:

• $m$ is the mass
• $\omega$ (omega) is the angular velocity in radians per second (rad/s)
• $r$ is the radius

First, we must convert the frequency from rev/min to rev/s:

Next, convert frequency to angular velocity ($\omega$) using the formula $\omega = 2\pi f$:

3. Show the solution

Now, substitute the values into the centripetal force formula:

Rounding to the nearest whole number, we get 379 N.

4. Identify the answer

Comparing our result with the options provided, the correct option is D.

5. Tip

When dealing with circular motion problems in JAMB, always ensure your angular velocity ($\omega$) is in radians per second. If the question gives you revolutions per minute, divide by 60 first to get Hertz, then multiply by $2\pi$. Also, remember that $\pi^2$ is approximately 9.87, which is often useful for quick calculations!