Bifilar Suspension Experiment

TO STUDY OSCILLATIONS IN BIFILAR SUSPENSION ARRANGEMENT

1. AIM

To determine the moment of inertia of an irregular lamina using bifilar suspension arrangement and to verify the theorem of parallel axes.

2. APPARATUS USED

  • Bifilar suspension setup (two parallel threads of equal length)
  • Rigid lamina (rectangular or circular)
  • Meter scale
  • Vernier caliper
  • Stopwatch
  • Set of weights
  • Graph paper
  • Calculator

3. DIAGRAM

Bifilar Suspension Arrangement
Figure : Bifilar Suspension Arrangement

4. THEORY

The bifilar suspension arrangement consists of a rigid body suspended by two parallel vertical threads or wires of equal length. When the body is given a small angular displacement about the vertical axis passing through its center of mass, it executes simple harmonic oscillations.

In this arrangement, when the lamina is twisted and released, it oscillates in a horizontal plane about a vertical axis. The restoring torque is provided by the tension in the threads.

For small oscillations, the time period (T) of oscillation is related to the moment of inertia (I) of the lamina about the vertical axis passing through its center of mass by the formula:

\[ T = 2\pi\sqrt{\frac{IL^2}{Mgd^2}} \]

Where:

  • I = Moment of inertia of the lamina about the vertical axis through its center of mass
  • L = Length of the suspending threads
  • M = Mass of the lamina
  • g = Acceleration due to gravity
  • d = Half the distance between the threads

According to the theorem of parallel axes:

\[ I = I_{CM} + Mh^2 \]

Where:

  • I = Moment of inertia about any axis
  • ICM = Moment of inertia about an axis passing through the center of mass
  • M = Mass of the body
  • h = Perpendicular distance between the two axes

5. FORMULA

1. Time period of oscillation:

\[ T = 2\pi\sqrt{\frac{IL^2}{Mgd^2}} \]

2. Moment of inertia about the vertical axis:

\[ I = \frac{T^2Mgd^2}{4\pi^2L} \]

3. Theoretical moment of inertia:

For a rectangular lamina (about center): \[ I_{CM} = \frac{M(a^2 + b^2)}{12} \]

For a circular disc (about center): \[ I_{CM} = \frac{MR^2}{2} \]

4. Verification of parallel axis theorem:

\[ I = I_{CM} + Mh^2 \]

6. PROCEDURE

  1. Measure the mass (M) of the lamina using a physical balance.
  2. Set up the bifilar suspension by attaching two threads of equal length to the support.
  3. Measure the length (L) of the suspending threads.
  4. Attach the lamina to the lower ends of the threads such that:
    • The threads are parallel to each other
    • The center of gravity of the lamina lies on the vertical line midway between the threads
  5. Measure the distance (2d) between the two threads.
  6. For the center of mass experiment:
    • Ensure the threads are attached at equal distances from the center of the lamina
    • Give a small angular displacement (< 5°) to the lamina about the vertical axis
    • Release it gently and measure the time for 20 complete oscillations
    • Calculate the time period (T) by dividing the total time by the number of oscillations
    • Repeat the observation three times and take the average value of T
  7. For parallel axis theorem verification:
    • Attach the threads at different points, maintaining equal distance from a parallel axis
    • Measure the distance (h) between the center of mass and the axis of rotation
    • Repeat steps 6b to 6e to find the new time period
  8. Calculate the moment of inertia using the formula in section 5.
  9. Compare the experimental value with the theoretical value.

7. OBSERVATION TABLE

Part A: Determination of Moment of Inertia about Center of Mass

Parameters Values
Mass of the lamina (M) _______ kg
Length of the threads (L) _______ m
Distance between threads (2d) _______ m
Half distance between threads (d) _______ m
Observation No. Time for 20 oscillations (s) Time period (T) (s)
1
2
3
Average

Part B: Verification of Parallel Axis Theorem

Distance from CM (h) Time for 20 oscillations (s) Time period (T) (s) Moment of Inertia (I) (kg·m²)
h₁ = ___ m
h₂ = ___ m
h₃ = ___ m

8. CALCULATIONS

Part A: Moment of Inertia about Center of Mass

  1. Calculate the average time period (T):

    \[ T = \frac{T_1 + T_2 + T_3}{3} \]

  2. Calculate the experimental moment of inertia (ICM):

    \[ I_{CM} = \frac{T^2Mgd^2}{4\pi^2L} \]

  3. Calculate the theoretical moment of inertia:

    For rectangular lamina: \[ I_{CM(theo)} = \frac{M(a^2 + b^2)}{12} \]

    For circular disc: \[ I_{CM(theo)} = \frac{MR^2}{2} \]

  4. Calculate the percentage error:

    \[ \% \text{ Error} = \left|\frac{I_{CM(theo)} - I_{CM(exp)}}{I_{CM(theo)}}\right| \times 100 \]

Part B: Verification of Parallel Axis Theorem

  1. For each value of h, calculate the moment of inertia (I) using the formula:

    \[ I = \frac{T^2Mgd^2}{4\pi^2L} \]

  2. Calculate the expected moment of inertia using parallel axis theorem:

    \[ I_{(theo)} = I_{CM} + Mh^2 \]

  3. Compare the experimental and theoretical values:

    \[ \% \text{ Error} = \left|\frac{I_{(theo)} - I_{(exp)}}{I_{(theo)}}\right| \times 100 \]

9. RESULT

  1. The moment of inertia of the lamina about its center of mass is _______ kg·m².
  2. The theoretical value of the moment of inertia is _______ kg·m².
  3. The percentage error in the experiment is _______%.
  4. The parallel axis theorem is verified with an average percentage error of _______%.

10. PRECAUTIONS

  1. The suspending threads should be of equal length.
  2. The threads must be parallel to each other.
  3. The angular displacement given to the lamina should be small (< 5°) to ensure simple harmonic motion.
  4. The center of mass of the lamina should lie on the vertical line midway between the two threads.
  5. The lamina should be released gently without giving it any linear velocity.
  6. The amplitude of oscillations should be kept small to minimize air resistance effects.
  7. Time measurement should start when the lamina passes through the mean position.
  8. Count oscillations carefully to avoid counting errors.
  9. Avoid any draft or air currents in the room during the experiment.
  10. The suspension points should be rigidly fixed to avoid any unwanted oscillations.

11. VIVA VOCE QUESTIONS

1. What is bifilar suspension?

Bifilar suspension is an arrangement where a rigid body is suspended by two parallel threads or wires of equal length. It is used to study rotational oscillations and measure moments of inertia.

2. Define moment of inertia and state its SI unit.

Moment of inertia is the rotational analog of mass in linear motion. It represents the resistance of a body to angular acceleration. Its SI unit is kg·m².

3. State the parallel axis theorem.

The parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia about a parallel axis passing through its center of mass and the product of its mass and the square of the perpendicular distance between the two axes.

4. How does the time period depend on the length of the threads?

The time period is directly proportional to the square root of the length of the threads. If the length is increased four times, the time period doubles.

5. Why should the angular displacement be kept small in this experiment?

Small angular displacements ensure that the restoring torque is proportional to the angular displacement, satisfying the condition for simple harmonic motion. Larger displacements introduce non-linearities in the restoring force.

6. What happens to the time period if the mass of the lamina is doubled?

If the mass is doubled while keeping all other parameters constant, the time period remains unchanged because both M in the numerator and M in the denominator cancel out in the formula T = 2π√(IL²/Mgd²).

7. How does the distance between the threads affect the time period?

The time period is inversely proportional to the distance between the threads. If the distance between the threads is doubled, the time period becomes half.

8. What are the sources of errors in this experiment?

Sources of error include non-uniform thickness of the lamina, air resistance, non-parallel threads, measurement errors in length and time, and imperfect release of the lamina.

9. Why is bifilar suspension preferred over torsional pendulum for measuring moment of inertia?

Bifilar suspension does not require knowledge of the torsional constant of the wire, which is difficult to determine accurately. It only requires measurements of easily measurable quantities like length, mass, and time period.

10. How would the time period change if the experiment is performed on the moon?

On the moon, the acceleration due to gravity is about 1/6th of that on Earth. Since T ∝ 1/√g, the time period would increase by a factor of √6 or about 2.45 times.

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