What is the linear speed of the hoop's center of mass as it leaves the incline and rolls onto the horizontal surface?
The linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto the horizontal surface is given by √(21.56 + 4r²).
Calculation Explanation:
The problem states that a 3.5 kg hoop starts from rest at a height of 1.1m above the base of an inclined plane and rolls down under the influence of gravity. We need to find the linear speed of the hoop's center of mass as it leaves the incline and rolls onto the horizontal surface.
The formula for the velocity of the center of mass of the hoop can be derived from the conservation of energy principle. The total energy of the hoop at height h is equal to the energy of the hoop at the bottom of the inclined plane and on the horizontal surface.
Using the formulas for potential energy (mgh) and kinetic energy (1/2mv^2), we can equate them to find the velocity v. The equation becomes mgh = 1/2mv^2 + 1/2Iω², where I = mk² and k = r for a hoop.
Substituting the values of mass, height, and acceleration due to gravity, we find that v = √(2gh + r²ω²). Given that the hoop starts from rest, the initial velocity u is 0.
Substituting the known values into the equation, we find v = √(2 × 9.8 × 1.1 + 1/2 × 3.5 × (2r)² × 9.8/2 × 3.5) = √(21.56 + 4r²).
Therefore, the linear speed of the hoop's center of mass just as it leaves the incline and rolls onto the horizontal surface is √(21.56 + 4r²).