Question

A bob of mass 0.1 \(\text {kg}\) hung from the ceiling of a room by a string 2 \(\text m\) long is set into oscillation. The speed of the bob at its mean position is 1 \(\text m\,\text s^{-1}\). What is the trajectory of the bob if the string is cut when the bob is 

1. at one of its extreme positions, 
2. at its mean position.

Answer 1

Key Concept

When string cut → tension disappears → only gravity acts → parabolic projectile motion

Trajectory depends on position and velocity at cut instant

● Ceiling
|
| L = 2 m
\ ● Bob (θ_max or mean)

String CUT → Parabolic path

Final Answers

• (a) Extreme position: Falls vertically downward
• (b) Mean position: Follows parabolic trajectory

(a) Extreme Position Analysis

At extreme: velocity = 0 (momentarily stops)

$$\vec{v} = 0 \, \text{m/s} \quad \text{(both components zero)}$$

Result: Pure free fall → straight down under gravity

(b) Mean Position Analysis

At mean position: maximum horizontal speed = 1 m/s

$$v_\text{horizontal} = 1 \, \text{m/s}, \quad v_\text{vertical} = 0$$

Projectile motion:

$$x(t) = (1)t, \quad y(t) = -( \frac{1}{2}gt^2)$$

Result: Parabolic trajectory (horizontal velocity + vertical free fall)

Energy Check (confirms v_max = 1 m/s)

$$mgL(1-\cos\theta_\text{max}) = \frac{1}{2}mv_\text{max}^2$$ $$v_\text{max} = \sqrt{2gL(1-\cos\theta_\text{max})} = 1 \, \text{m/s}$$

Visual Summary

Position	Velocity at Cut	Trajectory
Extreme	\(v_x = 0, v_y = 0\)	Vertical fall
Mean	\(v_x = 1, v_y = 0\)	Parabola

Physical Insight

• String provides centripetal force + tension component
• Cut string → only gravity remains
• Existing velocity determines horizontal motion
• Zero initial vertical velocity → pure parabolic path