A hoop starts from rest at a height 1.8 m above the base of an inclined plane and rolls down under the influence of gravity. 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? (Neglect friction.)
I don't know what equation to use, other equations online don't define the variables.
gravitational potential energy is converted to kinetic energy
m g h = 1/2 m v²
9.8 * 1.8 = v² / 2
Thank You this is what i needed
To determine 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. This principle states that the total mechanical energy of a system remains constant if no external forces are acting on it.
Here's how we can proceed:
1. Identify the initial and final positions of the hoop's center of mass:
- Initial position: The center of mass is at a height of 1.8 m above the base of the incline.
- Final position: The center of mass is on the horizontal surface.
2. Apply the conservation of energy principle:
Since no external forces (such as friction) are acting on the hoop, the total mechanical energy is conserved. The initial potential energy of the hoop is converted into its final kinetic energy.
The total mechanical energy (E) can be written as the sum of the potential energy (PE) and the kinetic energy (KE) of the hoop's center of mass:
E = PE + KE.
3. Determine the initial potential energy (PE):
The potential energy at the initial position is given by the formula:
PE = mgh,
where m is the mass of the hoop, g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical displacement (1.8 m).
4. Determine the final kinetic energy (KE):
The kinetic energy at the final position can be calculated using the formula:
KE = 0.5 * m * v^2,
where v is the linear speed of the hoop's center of mass.
5. Equate the initial potential energy to the final kinetic energy:
Since the total mechanical energy is conserved, we can set E = PE + KE = constant.
Thus, we have mgh = 0.5 * m * v^2.
6. Solve for the linear speed (v):
Rearrange the equation to solve for v:
v = sqrt((2 * g * h)),
where sqrt denotes the square root.
You can now use the formula v = sqrt((2 * g * h)) to calculate the linear speed of the hoop's center of mass as it leaves the incline and rolls onto the horizontal surface. Substitute the values for g (9.8 m/s^2) and h (1.8 m) into the equation and then calculate the value of v.