91Ó°ÊÓ

Centripetal force is the force that keeps an object moving in a circular path. It always acts towards the center of the circle. In the case of the ring rolling inside the hoop, the centripetal force is crucial as it keeps the ring from flying off at the highest point of its trajectory. When the ring is at the top, the normal force becomes zero, and the only force providing the centripetal acceleration is the gravitational force acting downwards.
The formula for centripetal force is given by \( F_c = \frac{mv^2}{R} \), where:

• \( F_c \) is the centripetal force
• \( m \) is the mass of the object
• \( v \) is the velocity of the object
• \( R \) is the radius of the circular path

In this problem, the gravitational force \( mg \) equals the centripetal force \( \frac{mv^2}{R} \). This equation allows us to solve for the linear velocity of the ring at the highest point.

Pure rolling, also known as rolling without slipping, occurs when there is no relative motion between the contact point of the rolling object and the surface it is rolling on. For the ring inside the hoop, pure rolling ensures a direct relationship between linear velocity and angular velocity.
In pure rolling, the condition \( v = R_{\text{ring}} \cdot \omega \) holds true, where:

• \( v \) is the linear velocity of the ring's center of mass
• \( R_{\text{ring}} \) is the radius of the ring
• \( \omega \) is the angular velocity

Knowing this relationship is key to determining the angular velocity of the ring when we have found its linear velocity. For our problem, this equation becomes \( v = \frac{R}{4} \cdot \omega \), allowing us to solve for \( \omega \).

Gravitational force is the attractive force between any two masses. On Earth, this force attracts objects towards the center of the planet, with a strength calculated by \( F_g = mg \), where:

• \( F_g \) is the gravitational force
• \( m \) is the mass of the object
• \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \) (rounded to \( 10 \, \text{m/s}^2 \) as given in this problem)

In this exercise, gravitational force plays a dual role. It acts as both the force pulling the ring downwards at every point in its path and as the necessary centripetal force at the highest point to keep the ring in motion along the spherical trajectory of the hoop.

Linear velocity refers to the speed at which an object moves along a straight path. It measures how fast the position of an object is changing with time. When dealing with circular motion and rolling objects, linear velocity is often associated with the speed of the center of mass.
In the context of pure rolling, linear velocity is directly connected to angular velocity by the radius of the rolling object. For the ring inside the hoop rolling without slipping, we use the relationship \( v = R_{\text{ring}} \cdot \omega \) to find linear velocity from angular velocity or vice versa.
In our problem, we calculate this velocity when the centric force is provided solely by gravity, resulting in \( v = \sqrt{gR} \). This step provides \( v = \sqrt{3} \), and using the radius of the ring, we further deduce the angular velocity.