Answer:
Explanation:
To calculate the formula for the speed of the center of mass of the cylinder at the bottom of the incline, we can use the principle of conservation of energy.
The initial potential energy of the cylinder at height h is given by mgh, where m is the mass and g is the acceleration due to gravity.
At the bottom of the incline, all of this potential energy is converted into kinetic energy. The kinetic energy can be split into two components: the kinetic energy of translation and the kinetic energy of rotation.
The kinetic energy of translation is given by (1/2)mv^2, where v is the speed of the center of mass.
The kinetic energy of rotation is given by (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.
For a solid cylinder rolling without slipping, the relationship between linear and angular velocity is v = Rω, where R is the radius of the cylinder.
The moment of inertia for a solid cylinder rotating about its axis (perpendicular to the plane of motion) is given by I = (1/2)mr^2.
Substituting the relationship between linear and angular velocity, we can express the kinetic energy of rotation as (1/2)(1/2)mr^2(v/R)^2 = (1/4)mv^2.
Equating the total initial potential energy to the sum of the kinetic energy of translation and the kinetic energy of rotation, we have:
mgh = (1/2)mv^2 + (1/4)mv^2
Simplifying, we get:
mgh = (3/4)mv^2
Cancelling out the mass m, we get:
gh = (3/4)v^2
Taking the square root of both sides, the formula for the speed of the center of mass (v) at the bottom of the incline is:
v = √(4gh/3)
Comparing this result to the speed of a block sliding down a frictionless plane from height h, Mock = √(2gh), we see that the speed of the cylinder is slightly higher. This is because a rolling object has both translational and rotational kinetic energy, while a sliding object only has translational kinetic energy.
To find the percentage of the total energy that is "diverted" to rolling kinetic energy for the cylinder, we can calculate the ratio of the kinetic energy of rotation to the total energy.
The total energy (E) is equal to the sum of the potential energy at height h and the kinetic energy at the bottom:
E = mgh + (1/2)mv^2 + (1/4)mv^2 = mgh + (3/4)mv^2
The fraction of the total energy that is due to the rotational kinetic energy (Kr) is given by:
Kr/E = (1/4)mv^2 / [mgh + (3/4)mv^2]
Cancelling out the mass m, we have:
Kr/E = (1/4)v^2 / [gh + (3/4)v^2]
To calculate the percentage, we multiply this fraction by 100:
Percentage = (1/4)v^2 / [gh + (3/4)v^2] * 100
Substituting the formula for v, we have:
Percentage = (1/4)(4gh/3) / [gh + (3/4)(4gh/3)] * 100
Simplifying, we get:
Percentage = 100/7 ≈ 14.29%
So, approximately 14.29% of the total energy is "diverted" to rolling kinetic energy for the cylinder.