Home
Class 11
PHYSICS
A sphere of mass 1 kg has a radius of 10...

A sphere of mass 1 kg has a radius of 10 cm. Calculate the moment of inertia (i) about the diameter and (ii) about the tangent.

Text Solution

AI Generated Solution

The correct Answer is:
To calculate the moment of inertia of a sphere about its diameter and about a tangent, we can follow these steps: ### Step 1: Calculate the Moment of Inertia about the Diameter The formula for the moment of inertia \( I \) of a solid sphere about an axis through its center (diameter) is given by: \[ I = \frac{2}{5} m R^2 \] Where: - \( m \) is the mass of the sphere - \( R \) is the radius of the sphere Given: - Mass \( m = 1 \, \text{kg} \) - Radius \( R = 10 \, \text{cm} = 0.1 \, \text{m} \) Substituting the values into the formula: \[ I = \frac{2}{5} \times 1 \, \text{kg} \times (0.1 \, \text{m})^2 \] Calculating this: \[ I = \frac{2}{5} \times 1 \times 0.01 = \frac{2}{500} \, \text{kg m}^2 = 0.004 \, \text{kg m}^2 \] ### Step 2: Calculate the Moment of Inertia about the Tangent To find the moment of inertia about a tangent, we can use the Parallel Axis Theorem, which states: \[ I = I_{\text{cm}} + m d^2 \] Where: - \( I_{\text{cm}} \) is the moment of inertia about the center of mass (which we calculated in Step 1) - \( d \) is the distance from the center of mass to the new axis (which is equal to the radius \( R \) of the sphere) Substituting the values: \[ I_{\text{tangent}} = I_{\text{cm}} + m R^2 \] \[ I_{\text{tangent}} = \frac{2}{5} m R^2 + m R^2 \] Factoring out \( m R^2 \): \[ I_{\text{tangent}} = m R^2 \left( \frac{2}{5} + 1 \right) = m R^2 \left( \frac{2}{5} + \frac{5}{5} \right) = m R^2 \left( \frac{7}{5} \right) \] Now substituting \( m = 1 \, \text{kg} \) and \( R = 0.1 \, \text{m} \): \[ I_{\text{tangent}} = 1 \, \text{kg} \times (0.1 \, \text{m})^2 \times \frac{7}{5} \] \[ I_{\text{tangent}} = 1 \times 0.01 \times \frac{7}{5} = \frac{7}{500} \, \text{kg m}^2 = 0.014 \, \text{kg m}^2 \] ### Final Answers: (i) Moment of inertia about the diameter: \( 0.004 \, \text{kg m}^2 \) (ii) Moment of inertia about the tangent: \( 0.014 \, \text{kg m}^2 \) ---
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ROTATIONAL MOTION

    SL ARORA|Exercise Problem|24 Videos
  • PROPERTIES OF SOLIDS

    SL ARORA|Exercise All Questions|97 Videos
  • System of particles & rotational Motion

    SL ARORA|Exercise EXERCISE|353 Videos

Similar Questions

Explore conceptually related problems

The radius of a sphere is 5 cm. Calculate the radius of gyration about (i) its diameter and (ii) about any tangent.

A sphere of mass 10 kg and radius 0.5 m rotates about a tangent. The moment of inertia of the solid sphere about tangent is

Knowledge Check

  • A sphere of mass 10 kg and radius 0.5 m rotates about a tangent. The moment of inertia of the sphere is

    A
    5 kg `m^(2)`
    B
    `2.7` kg `m^(2)`
    C
    3.5 kg `m^(2)`
    D
    4.5 kg `m^(2)`
  • A sphere of mass 10 kg and radius 0.5 m rotates about a tangent. The moment of inertia of the sphere is

    A
    `5 kg m^(2)`
    B
    `2.7 kg m^(2)`
    C
    `3.5 kg m^(2)`
    D
    `4.5 kg m^(2)`
  • The moment of inertia of a solid sphere of mass M and radius R about its diameter is I. The moment of inertia of the same sphere about a tangent parallel to the diameter is

    A
    5 I
    B
    7 I
    C
    2.5 I
    D
    3.5 I
  • Similar Questions

    Explore conceptually related problems

    What is moment of inertia of a solid sphere about its diameter ?

    A cylinder of 500 g and radius 10 cm has moment of inertia (about its natural axis)

    The radius of a sphere is 8 cm. What will be its radius of gyration about (i) its diameter and (ii) about any tangent?

    If the radius of gyration of a solid disc of mass 10 kg about an axis is 0.40 m, then the moment of inertia of the disc about that axis is

    Two spheres of same mass and radius are in contact with each other. If the moment of inertia of a sphere about its diameter is I, then the moment of inertia of both the spheres about the tangent at their common point would be