Question

A particle moves in a horizontal circle. If its speed is doubled, the centripetal force acting on it becomes:

• A. Same
• B. Double
• C. Four times
• D. Half

Hint:

Because the force depends on the *square* of the velocity ($F \propto v^2$), any scaling factor applied to the velocity is squared when determining the new force: $$\text{Speed } \times 2 \implies \text{Force } \times 2^2 = 4$$ $$\text{Speed } \times 3 \implies \text{Force } \times 3^2 = 9$$

Answer 1

The Correct Option is C

Step 1: Understand the setup.
A particle moves in a horizontal circle. To keep moving in a circle it needs an inward pull called the centripetal force. We want to see how this force changes when the speed is doubled.

Step 2: Write the centripetal force formula.
The inward force needed for circular motion is: \[ F = \frac{m v^2}{r} \] Here $m$ is the mass, $v$ is the speed, and $r$ is the radius of the circle.

Step 3: Note what stays fixed.
The mass of the particle does not change. The radius of the circle also stays the same. Only the speed is changed. So the force depends only on the square of the speed here.

Step 4: Double the speed.
Replace $v$ with $2v$ in the formula: \[ F_{new} = \frac{m (2v)^2}{r} = \frac{m \cdot 4 v^2}{r} \]

Step 5: Compare with the old force.
We can pull out the factor of $4$: \[ F_{new} = 4 \times \frac{m v^2}{r} = 4F \] So the new force is four times the original force.

Step 6: State the result.
Because force depends on speed squared, doubling the speed makes the force grow by a factor of $2^2 = 4$. The centripetal force becomes four times larger. \[ \boxed{\text{Four times}} \]