The kinetic energy K of a particle moving along a circle of radius 2 metre depends upon the distance covered d as K= cd^2 . the ratio of tangential and centripetal acceleration is

To find the ratio of tangential and centripetal acceleration for a particle moving along a circular path, we first need to understand how kinetic energy relates to these types of acceleration. Given that the kinetic energy \( K \) of the particle is expressed as \( K = cd^2 \), where \( c \) is a constant and \( d \) is the distance covered, we can derive the necessary components step by step.

Understanding Kinetic Energy and Velocity

The kinetic energy of an object is related to its velocity by the formula:

K = \frac{1}{2} mv^2

Here, \( m \) is the mass of the particle, and \( v \) is its velocity. Since we have \( K = cd^2 \), we can equate the two expressions for kinetic energy:

cd^2 = \frac{1}{2} mv^2

Finding Velocity

From the equation above, we can solve for the velocity \( v \):

v^2 = \frac{2cd^2}{m}

Taking the square root gives us:

v = \sqrt{\frac{2cd^2}{m}}

Calculating Tangential Acceleration

The tangential acceleration \( a_t \) is defined as the rate of change of velocity with respect to time. Since the velocity depends on the distance covered \( d \), we can express tangential acceleration as:

a_t = \frac{dv}{dt} = \frac{dv}{dd} \cdot \frac{dd}{dt} = \frac{dv}{dd} \cdot v

To find \( \frac{dv}{dd} \), we differentiate \( v \) with respect to \( d \):

\(\frac{dv}{dd} = \frac{d}{dd}\left(\sqrt{\frac{2cd^2}{m}}\right) = \frac{2\sqrt{\frac{2c}{m}} \cdot d}{2\sqrt{2cd^2}} = \sqrt{\frac{2c}{m}}\)

Thus, the tangential acceleration becomes:

a_t = \sqrt{\frac{2c}{m}} \cdot \sqrt{\frac{2cd^2}{m}} = \frac{2c}{m}

Calculating Centripetal Acceleration

Centripetal acceleration \( a_c \) is given by the formula:

a_c = \frac{v^2}{r}

Substituting our expression for \( v^2 \):

a_c = \frac{\frac{2cd^2}{m}}{r}

Given that the radius \( r \) is 2 meters, we have:

a_c = \frac{2cd^2}{2m} = \frac{cd^2}{m}

Finding the Ratio of Accelerations

Now that we have both tangential and centripetal accelerations, we can find their ratio:

\(\frac{a_t}{a_c} = \frac{\frac{2c}{m}}{\frac{cd^2}{m}} = \frac{2}{d^2}\)

Final Thoughts

The ratio of tangential to centripetal acceleration for the particle moving along a circular path is therefore:

\(\frac{a_t}{a_c} = \frac{2}{d^2}\)

This ratio indicates how the tangential acceleration changes relative to the centripetal acceleration as the distance \( d \) varies. As \( d \) increases, the centripetal acceleration increases more significantly than the tangential acceleration, which is an important consideration in circular motion dynamics.