A solid sphere starts from rest at the upper end of the track and rolls without slipping until it rolls off the right-hand end. If H=12.0 m and h=6.0 m and the track is horizontal at the right-hand end, how far horizontally from point A does the sphere land on the floor?
To find the horizontal distance from point A where the sphere lands on the floor, we can use the principle of conservation of energy.
Let's consider the initial and final states of the sphere. At the beginning, the sphere is at point A, at height H. At the end, it lands on the floor at an unknown horizontal distance from point A.
The initial energy of the sphere is purely potential energy, given by mgh, where m is the mass of the sphere, g is the acceleration due to gravity, and h is the initial height. The final energy of the sphere is the sum of potential energy and kinetic energy.
Since the sphere is rolling without slipping, its final kinetic energy is a combination of translational and rotational kinetic energy. The translational kinetic energy is given by 1/2mv^2, where v is the velocity of the sphere at the bottom of the track. The rotational kinetic energy is given by 1/2Iω^2, where I is the moment of inertia of the sphere and ω is its angular velocity.
Since the sphere rolls without slipping, we have v = ωR, where R is the radius of the sphere. Substituting this relation into the expressions for translational and rotational kinetic energy, we get 1/2mv^2 = 1/2I(v/R)^2.
Equating the initial and final energies, we have mgh = 1/2mv^2 + 1/2I(v/R)^2.
The moment of inertia of a solid sphere about its diameter is (2/5)mR^2. Substituting this value into the equation, we get mgh = 1/2mv^2 + 1/2(2/5)m(v/R)^2.
Canceling out the common factors of m and multiplying both sides by 2/5, we get 2/5gh = 1/2v^2 + 1/5(v/R)^2.
The velocity v at the bottom of the track can be found using the principle of conservation of energy. The potential energy at height H is converted into kinetic energy at the bottom of the track. Thus, mgh = 1/2mv^2. Canceling out the common factors of m, we get gh = 1/2v^2.
Substituting this expression for gh in the previous equation, we have 2/5(1/2v^2) = 1/2v^2 + 1/5(v/R)^2.
Simplifying the equation, we get 1/5v^2 = 1/5(v/R)^2, which can be further simplified to v^2 = (v/R)^2.
Taking the square root of both sides, we get v = v/R.
Canceling out the common factor of v, we have 1 = 1/R.
Therefore, R = 1.
Since the track is horizontal at the right-hand end, the horizontal distance from point A where the sphere lands on the floor is equal to the distance it rolls along the track.
Since the sphere is rolling without slipping, the distance it rolls is equal to its circumference. The circumference of a sphere is given by 2πR, where R is the radius of the sphere.
Substituting the value of R as 1, we have the horizontal distance = 2π(1) = 2π.
Therefore, the sphere lands on the floor at a horizontal distance of 2π meters from point A.