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Q5.12(a):
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is $1 m s^{-1}$. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions,

Solution :

Step 1: Analyzing the Kinematic State at the Extreme Position

The bob is undergoing simple harmonic motion (pendulum oscillation). At the extreme position of its trajectory, the velocity ($v$) of the bob is momentarily zero ($v = 0 \, \text{m s}^{-1}$) [Per the principles of Simple Harmonic Motion, kinetic energy is zero at maximum displacement].

The bob is at a vertical distance $h$ below the ceiling. Using the geometry of the pendulum, the vertical height is defined by $h = L \cos(\theta)$, where $L = 2 \, \text{m}$. Since the bob is held by the string, it is currently in a state of static equilibrium regarding its velocity but under tension.

Step 2: Force Analysis upon Severing the String

At the instant the string is cut at the extreme position, the following conditions apply:

  • Initial Velocity ($u$): $0 \, \text{m s}^{-1}$
  • Acting Forces: Only the gravitational force ($F_g = mg$) acts on the bob [Per Newton's Second Law of Motion].
  • Acceleration: The bob experiences a constant downward acceleration $a = g \approx 9.8 \, \text{m s}^{-2}$.

Step 3: Determining the Trajectory

Since the initial horizontal velocity ($v_x$) is $0$ and the initial vertical velocity ($v_y$) is $0$, the bob does not possess any horizontal component of motion. The body is subject only to the acceleration due to gravity acting vertically downward.

Bob (At extreme) Trajectory: Free Fall

Because the velocity vector is initially zero, the object does not follow a parabolic path. It falls vertically downward starting from the point of release [Per the equations of motion under gravity where $s = ut + \frac{1}{2}gt^2$ and $v_x = 0$].

Step 4: Conclusion of Motion

As there is no horizontal impetus, the bob will simply fall along the vertical line that passes through its position at the extreme point.

Final Solution: The trajectory of the bob is a vertical straight line directed downwards.


More Questions from Class 11 Physics Laws of Motion EXERCISES


CBSE Solutions for Class 11 Physics Laws of Motion


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