Linear Speed of a Hoop Rolling Down an Inclined Plane
What is the linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface?
Final answer:
The linear speed of the hoop's center of mass just as it leaves the incline and rolls onto a horizontal surface is approximately 4.43 m/s.
Answer:
To find the linear speed of the hoop's center of mass just as it leaves the incline and rolls onto a horizontal surface, we can use the principle of conservation of energy. The hoop starts from rest at a height of 1.00 m above the base of the inclined plane, so it possesses potential energy which will be converted into kinetic energy as it rolls down.
The potential energy at the initial height is given by mgh, where m is the mass of the hoop, g is the acceleration due to gravity, and h is the height. Since there is no friction, the total mechanical energy of the system is conserved. At the bottom of the incline, all potential energy is converted into kinetic energy.
Therefore, we can equate the potential energy to the kinetic energy: mgh = 1/2 mv^2. Solving for v, we get: sqrt(2gh). Plugging in the given values, we have: sqrt(2*9.81*1.00) = 4.43 m/s.
So, the linear speed of the hoop's center of mass just as it leaves the incline and rolls onto a horizontal surface is approximately 4.43 m/s.
When a 4.90-kilogram hoop starts from rest at a height of 1.00 m above the base of an inclined plane and rolls down under the influence of gravity, we can determine the linear speed of its center of mass as it leaves the incline and rolls onto a horizontal surface.
By applying the principle of conservation of energy and the conversion of potential energy to kinetic energy, we can calculate that the linear speed of the hoop's center of mass is approximately 4.43 m/s at that moment.
Understanding this concept is crucial in mechanics and physics to analyze the motion of objects in different scenarios and environments.