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

Concept: For a particle of mass $m$ executing uniform circular motion along a path of constant radius $r$ at a linear tangential speed $v$, it experiences a continuous inward radial acceleration called centripetal acceleration. According to Newton's Second Law of Motion, this requires a net inward centripetal force ($F_c$), given by: $$F_c = \frac{mv^2}{r}$$ This indicates that for a path of fixed radius, the centripetal force is directly proportional to the square of the linear speed of the particle: $$F_c \propto v^2$$

Step 1: Let the initial speed of the particle be $v_1 = v$, and the corresponding initial centripetal force be: $$F_1 = \frac{mv^2}{r}$$ According to the conditions specified in the problem, the speed of the particle is doubled while keeping the mass $m$ and the circular path radius $r$ constant: $$\text{New speed } v_2 = 2v$$

Step 2: Substitute the value of $v_2$ into the force equation to determine the new centripetal force $F_2$: $$F_2 = \frac{m(v_2)^2}{r} = \frac{m(2v)^2}{r}$$ $$F_2 = \frac{m \cdot 4v^2}{r} = 4 \cdot \left(\frac{mv^2}{r}\right)$$

Step 3: Relating $F_2$ back to our original baseline force $F_1$: $$F_2 = 4 \cdot F_1$$ Hence, the centripetal force increases by a factor of four, which matches Option (C).