Final answer:
The acceleration of the center of mass of the hoop as it rolls down the ramp is given by a = g * sin(theta), the speed of the center of mass of the hoop when it reaches the bottom of the ramp is v = √(2 * g * sin(theta) * L * R), the horizontal distance from the edge of the table to where the hoop lands on the floor is d = (2 * g * sin(theta) * L * R) / √(2 * g * sin(theta) * L * R / (2 * h)), and the distance determined for the hoop will be greater than the distance determined for the disk.
Step-by-step explanation:
(a) To derive an expression for the acceleration of the center of mass of the hoop as it rolls down the ramp, we can use the principles of rotational and translational motion. The net force on the hoop is due to the component of the gravitational force that acts parallel to the ramp. This force can be calculated as F = m * g * sin(theta), where m is the mass of the hoop, g is the acceleration due to gravity, and theta is the angle of the ramp. The net torque on the hoop is zero since it rolls without slipping, so the equation for torque can be written as I * alpha = 0, where I is the moment of inertia of the hoop and alpha is the angular acceleration. Since the hoop rolls without slipping, we can relate the angular acceleration and the linear acceleration of the center of mass as alpha = a / R, where R is the radius of the hoop. Solving these equations, we get the expression for the linear acceleration of the center of mass of the hoop as a = g * sin(theta).
(b) To derive the expression for the speed of the center of mass of the hoop when it reaches the bottom of the ramp, we can use the principle of conservation of mechanical energy. At the top of the ramp, the hoop has gravitational potential energy equal to m * g * L, where L is the height of the ramp above the floor. At the bottom of the ramp, the hoop has kinetic energy equal to (1/2) * I * w², where w is the angular velocity of the hoop. Assuming no energy loss due to friction or other factors, we can equate these two expressions to get m * g * L = (1/2) * I * w². Since the hoop rolls without slipping, we can relate the linear velocity of the center of mass and the angular velocity as v = w * R. Solving for w, we get the expression for the angular velocity as w = √(2 * g * sin(theta) * L / R). Substituting this into the equation for the linear velocity, we get the expression for the speed of the center of mass of the hoop as v = √(2 * g * sin(theta) * L * R).
(c) To derive an expression for the horizontal distance from the edge of the table to where the hoop lands on the floor, we can consider the horizontal motion of the center of mass of the hoop after it reaches the bottom of the ramp. Assuming no horizontal forces act on the hoop, the horizontal distance traveled can be calculated using the equation d = v * t, where d is the distance, v is the speed of the center of mass, and t is the time. The time can be calculated as t = (2 * h) / v, where h is the height of the table above the floor. Substituting this into the equation for distance, we get the expression for the horizontal distance as d = (2 * g * sin(theta) * L * R) / √(2 * g * sin(theta) * L * R / (2 * h)).
(d) If the hoop is replaced by a disk having the same mass and radius, the distance from the edge of the table to where the disk lands on the floor will be greater than the distance determined for the hoop. This is because the moment of inertia of the hoop is greater than the moment of inertia of the disk for the same mass and radius. The moment of inertia affects the rotational motion of the object, specifically the angular acceleration and the speed of the center of mass. Since the disk has a smaller moment of inertia, it will accelerate more quickly and reach a greater speed, resulting in a longer horizontal distance traveled.