A UNIFORM RING OF MASS M AND RADIUS R IS ROLLING WITHOUT SLIPPING ON A HORIZONTAL SURFACE.FIND THE RADIUS OF CURVATURE OF TRAJECTORY TRACED OUT BY THE HIGHEST POINT?

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To find the radius of curvature of the trajectory traced out by the highest point of a rolling uniform ring, we can use the concept of centripetal acceleration.

When a ring rolls without slipping, both its rotational and translational motion are linked. The highest point on the trajectory has a net acceleration that can be divided into two components:

1. Tangential acceleration (a_t): The linear acceleration of the ring's center of mass along the curved path.
2. Centripetal acceleration (a_c): The acceleration toward the center of the circular path.

Since the ring rolls without slipping, the velocity of the point in contact with the ground is zero. At the highest point, the contact point has zero velocity, but the ring's center of mass has non-zero velocity due to its translational and rotational motion.

To find the radius of curvature, we need to find the relation between the tangential and centripetal accelerations:

a_t = α * R (Equation 1)
a_c = v^2 / R (Equation 2)

Where:
- α is the angular acceleration of the ring.
- R is the radius of curvature of the trajectory.
- v is the velocity of the center of mass.

Since the ring rolls without slipping, we have the relation between the linear and angular velocities:

v = ω * R (Equation 3)

Where:
- ω is the angular velocity of the ring.

Combining Equations 2 and 3, we can express the centripetal acceleration solely in terms of the angular velocity:

a_c = (ω * R)^2 / R
a_c = ω^2 * R

Now, let's find the relation between the angular acceleration and angular velocity:

a_t = α * R
α * R = a_t
α = a_t / R

Substituting the value of α in the centripetal acceleration equation:

a_c = (a_t / R) * R
a_c = a_t

Since the net acceleration at the highest point (a_net) is the vector sum of the tangential and centripetal accelerations:

a_net = √(a_t^2 + a_c^2)
a_net = √(a_t^2 + a_t^2)
a_net = √2 * a_t

At the highest point, a_net is equal to the acceleration due to gravity (g). Therefore:

g = √2 * a_t
a_t = g / √2

Finally, the radius of curvature (R) can be calculated by substituting the values of a_t, v, ω, and g in the equation (Equation 3):

R = v^2 / a_c
R = (ω * R)^2 / (g / √2)
R = ω^2 * R * √2 / g

Dividing both sides of the equation by R:

1 = ω^2 * √2 / g

Solving for R:

R = g / (ω^2 * √2)

Therefore, the radius of curvature of the trajectory traced out by the highest point of the rolling uniform ring is given by R = g / (ω^2 * √2), where g is the acceleration due to gravity and ω is the angular velocity of the ring.