Question

Bob of a simple pendulum of length $L$ is projected horizontally with a speed of $u = \sqrt{4gL}$, from the lowest position. Find the distance of the bob from vertical line $AB$, at the moment its tangential acceleration becomes zero.

Answer 1

0

Answer 2

The tangential acceleration ($a_t$) of a simple pendulum is given by $a_t = -g \sin\theta$, where $\theta$ is the angle the string makes with the downward vertical.

For $a_t$ to be zero, $\sin\theta$ must be zero, which means $\theta = 0$ or $\theta = \pi$.

1. At $\theta = 0$ (the lowest position): This is the initial point where the bob is projected. The tangential acceleration is indeed zero here. The horizontal distance from the vertical line AB is $L \sin 0 = 0$.
2. At $\theta = \pi$ (the highest point, vertically above the pivot): The initial speed $u = \sqrt{4gL}$ is such that the bob would just reach this point with zero speed if it were a rigid rod. However, for a string, the tension must remain non-negative. The tension in the string is $T = mg(3\cos\theta + 2)$. The string becomes slack when $T=0$, which occurs at $\cos\theta = -2/3$. This angle is approximately $131.8^\circ$, which is less than $180^\circ$ ($\pi$). Therefore, the bob does not reach $\theta = \pi$ while still being a pendulum.

Once the string becomes slack, the bob undergoes projectile motion. In projectile motion, the acceleration is purely gravitational ($g$ downwards). The tangential acceleration for projectile motion is $a_t = g \frac{v_y}{v}$, where $v_y$ is the vertical component of velocity. For $a_t=0$, $v_y$ must be zero, which corresponds to the highest point of the parabolic trajectory. At the point where the string becomes slack ($\theta = \arccos(-2/3)$), the bob's vertical velocity component is negative (it is already moving downwards). This means it has already passed the highest point of its projectile trajectory, or it will not reach a point where $v_y=0$ in its forward motion.

Thus, the only point where the tangential acceleration becomes zero is the initial lowest position. The horizontal distance of the bob from the vertical line AB at this point is $0$.