Question

A body attached to one end of a string performs motion along a vertical circle. Its centripetal acceleration, when the string is horizontal, will be [$g$ = acceleration due to gravity]

• A. zero
• B. $g$
• C. $3g$
• D. $5g$

Hint:

Memorize the velocity benchmarks for critical vertical loops: bottom is $\sqrt{5gr}$, horizontal is $\sqrt{3gr}$, and top is $\sqrt{gr}$. Because centripetal acceleration is simply $\frac{v^2}{r}$, dropping the square root and dividing by $r$ gives the acceleration values instantly: $5g$ at the bottom, $3g$ at the horizontal, and $g$ at the top.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
An object tied to a string of radius $r$ executes a standard vertical circular motion path under the continuous influence of gravity.
We need to calculate the exact magnitude of its centripetal acceleration ($a_c$) at the specific instant the string transitions through the perfectly horizontal position.

Step 2: Key Formula or Approach:
1. The centripetal acceleration of any object traversing a circular path of radius $r$ with instantaneous linear velocity $v$ is given by:
$$a_c = \frac{v^2}{r}$$ 2. According to the conservation of mechanical energy for a vertical circle, the minimum velocity at the horizontal position necessary to maintain the loop satisfies:
$$v^2 = 3gr$$

Step 3: Detailed Explanation:
Let's evaluate the mechanical energy transformation relative to the lowest point of the circle (the bottom position), where the velocity is $v_{bottom} = \sqrt{5gr}$.
As the mass ascends from the bottom to the horizontal midpoint level, it climbs a vertical height equal to the radius ($h = r$).
Using the conservation of mechanical energy:
$$\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv^2 + mgh$$ $$\frac{1}{2}m(5gr) = \frac{1}{2}mv^2 + mgr$$ Multiply the entire expression by $\frac{2}{m}$ to isolate the velocity terms:
$$5gr = v^2 + 2gr \implies v^2 = 3gr$$ Now, substitute this horizontal velocity expression directly into the standard definition for centripetal acceleration:
$$a_c = \frac{v^2}{r} = \frac{3gr}{r} = 3g$$

Step 4: Final Answer:
The centripetal acceleration when the string is horizontal is $3g$, matching option (C).