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a 25 kg uniform solid sphere with a 20 cm radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. a torque of 0.10 N-m is required to rotate the sphere through an angle of 1.0 rad and then the orientation is maintained.if the sphere is released then its time period of the oscilaation will be

a- sec c- sec

b- sec d- sec

the correct answer is d .

plz tell me how should we solve this problem.

Kevin Nash , 11 Years ago
Grade upto college level
anser 1 Answers
Askiitians Tutor Team

To solve the problem of the oscillation period of the suspended solid sphere, we need to understand the concepts of torque, moment of inertia, and simple harmonic motion. Let's break this down step by step.

Understanding the System

We have a solid sphere with a mass of 25 kg and a radius of 20 cm (0.2 m). The sphere is suspended by a wire, and when it is rotated through an angle of 1 radian, a torque of 0.10 N-m is applied. When released, the sphere will oscillate about its equilibrium position.

Calculating the Moment of Inertia

The moment of inertia (I) for a solid sphere about an axis through its center is given by the formula:

  • I = (2/5) * m * r²

Substituting the values:

  • m = 25 kg
  • r = 0.2 m

Now, calculate:

  • I = (2/5) * 25 kg * (0.2 m)²
  • I = (2/5) * 25 * 0.04
  • I = (2/5) * 1 = 0.4 kg·m²

Finding the Angular Frequency

The torque (τ) is related to the angular displacement (θ) and the moment of inertia (I) by the equation:

  • τ = I * α

Where α is the angular acceleration. Rearranging gives us:

  • α = τ / I

Substituting the known values:

  • α = 0.10 N-m / 0.4 kg·m²
  • α = 0.25 rad/s²

Calculating the Time Period of Oscillation

For small oscillations, the time period (T) of a physical pendulum can be calculated using the formula:

  • T = 2π * √(I / (m * g * d))

Where:

  • g = acceleration due to gravity (approximately 9.81 m/s²)
  • d = distance from the pivot point to the center of mass. For a sphere, this is equal to the radius (0.2 m).

Now substituting the values:

  • T = 2π * √(0.4 kg·m² / (25 kg * 9.81 m/s² * 0.2 m))

Calculating the denominator:

  • 25 kg * 9.81 m/s² * 0.2 m = 49.025 kg·m²/s²

Now, substituting back into the equation for T:

  • T = 2π * √(0.4 / 49.025)
  • T = 2π * √(0.00815)
  • T = 2π * 0.0904
  • T ≈ 0.568 sec

Final Result

Based on the calculations, the time period of oscillation is approximately 0.568 seconds. If the answer choices provided include this value, then it would correspond to option d. Thus, the correct answer is confirmed as d.

In summary, we calculated the moment of inertia, determined the angular acceleration from the torque, and finally used the physical pendulum formula to find the time period of oscillation. This systematic approach allows us to understand the dynamics of the suspended sphere clearly.

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