Calculate the radius of gyrations of a uniform metre rule suspended from 10cm mark oscillating as a compound pendulum. (Take mass m=50*10^_3 and l=4.2*10^_4)

To calculate the radius of gyration of a uniform metre rule suspended from the 10 cm mark and oscillating as a compound pendulum, we need to understand a few key concepts related to the motion of pendulums and the properties of rigid bodies. The radius of gyration is a measure that helps us understand how the mass of an object is distributed relative to an axis of rotation. It is denoted by the symbol 'k' and is defined as the distance from the axis of rotation at which the entire mass of the body can be assumed to be concentrated for the purpose of calculating rotational inertia.

Understanding the Setup

In this scenario, we have a uniform metre rule, which means its mass is evenly distributed along its length. The rule is suspended at the 10 cm mark, meaning that 90 cm of the rule hangs below the pivot point. The total length of the metre rule is 1 m (or 100 cm), and we are given the mass (m) as 50 grams (or 50 x 10^-3 kg) and the length (l) as 4.2 x 10^-4 m.

Calculating the Moment of Inertia

The moment of inertia (I) of a uniform rod about an axis through one end is given by the formula:

• I = (1/3) * m * l²

However, since the rod is pivoted at the 10 cm mark, we need to adjust our calculations. The length of the rod below the pivot is 90 cm (or 0.9 m), so we will use this length in our calculations. The moment of inertia about the pivot point can be calculated using the parallel axis theorem:

• I = I_cm + m * d²

Where:

• I_cm is the moment of inertia about the center of mass (which is at the midpoint of the rod),
• d is the distance from the center of mass to the pivot point.

Finding the Center of Mass and Distance

The center of mass of the metre rule is at the 50 cm mark. The distance (d) from the center of mass to the pivot point (10 cm mark) is:

• d = 50 cm - 10 cm = 40 cm = 0.4 m

Now, we can calculate I_cm for the entire rod:

• I_cm = (1/3) * m * l² = (1/3) * (50 x 10^-3 kg) * (1 m)² = (1/3) * 0.05 kg * 1 m² = 0.01667 kg·m²

Now, substituting into the parallel axis theorem:

• I = I_cm + m * d² = 0.01667 kg·m² + (50 x 10^-3 kg) * (0.4 m)²

Calculating the second term:

• (50 x 10^-3 kg) * (0.4 m)² = 0.008 kg·m²

Thus, the total moment of inertia is:

• I = 0.01667 kg·m² + 0.008 kg·m² = 0.02467 kg·m²

Calculating the Radius of Gyration

The radius of gyration (k) can be found using the formula:

• k = √(I/m)

Substituting the values we have:

• k = √(0.02467 kg·m² / (50 x 10^-3 kg))

Calculating this gives:

• k = √(0.4934 m²) ≈ 0.703 m

Final Result

Therefore, the radius of gyration of the uniform metre rule suspended from the 10 cm mark is approximately 0.703 meters. This value indicates how the mass is distributed relative to the pivot point, which is crucial for understanding the dynamics of the pendulum's oscillation.