A uniform circular disc of radius R oscillates in a vertical plane about a horizontal Axis. find the distance of the axis of rotation from the centre for which the period is minimum. what is the value of this period?

Question

Answer

Uniform Circular Disc Oscillation

In a uniform circular disc oscillation, the distance of the axis of rotation from the centre plays a crucial role in determining the period of the oscillation. Let us find out the distance at which the period is minimum and the corresponding value of the period.

Period of Oscillation

The period of oscillation of a uniform circular disc of radius R is given by:

T = 2π√(I/mgR)

where I is the moment of inertia of the disc, m is its mass, g is the acceleration due to gravity, and R is the radius of the disc.

Determining the Distance of the Axis of Rotation

Let us assume that the distance of the axis of rotation from the centre of the disc is x. Then, the moment of inertia of the disc about the axis of rotation is given by:

I = (1/2)mR^2 + (1/4)m(x^2 + R^2)

Substituting this value of I in the expression for T, we get:

T = 2π√[(1/2)R^2 + (1/4)(x^2 + R^2)]/(gR)

To find the value of x for which T is minimum, we differentiate T with respect to x and equate it to zero:

dT/dx = πx/2√[(1/2)R^2 + (1/4)(x^2 + R^2)]^3 = 0

Solving this equation, we get:

x = R/√2

Value of the Period

Substituting x = R/√2 in the expression for T, we get:

T = 2π√(3/4)R/g

This is the minimum value of the period of oscillation of the uniform circular disc. Its numerical value can be calculated using the given values of R and g.

Conclusion

Thus, we have determined the distance of the axis of rotation from the centre of the uniform circular disc for which the period of oscillation is minimum, and also found the corresponding value of the period.

Answer

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