How to Calculate the Linear Speed of a Hoop's Center of Mass
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?
The linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface can be calculated using the equation for conservation of energy. What is the linear speed of the hoop's center?
Answer:
The linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface is 5.41 m/s.
When calculating the linear speed of the hoop's center of mass, we can use the conservation of energy equation. The equation states that the initial energy of the system is equal to the final energy of the system.
In this case, the initial potential energy of the hoop is calculated by multiplying the mass of the hoop (3.2 kg) by the acceleration due to gravity (9.8 m/s²) by the height of the incline (1.40 m), which equals 44.56 J.
Since the hoop starts from rest, the initial kinetic energy is zero. Therefore, the conservation of energy equation can be written as KEf = 0 + 44.56 J.
The kinetic energy of the hoop just as it leaves the incline can be calculated by KEf = 3.2 kg × (v²) / 2, where v is the linear speed of the hoop's center of mass.
By combining the equations and solving for v, we find that the linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface is 5.41 m/s.