Can someone solve a physics problem concerning rotational motion for me?

Question

The problem is: In a circus performance, a large 5 kg hoop of radius 3 m rolls without slipping. If the hoop is given an angular speed of 3 rad/s while rolling on the horizontal and allowed to roll up a ramp inclined at 20 degrees with the horizontal, how far (measured along the incline) does the hoop roll?

Answer 1

The total energy of the hoop as it enters the ramp is the sum of its forward kinetic energy and its rotational energy. If it is rolling at w rad/sec, it is moving forward at a speed of 2*π*R*w. The total kinetic energy is then .5*M*[2*π*R*w]^2 + .5*I*w^2 I is the moment of inertia of the hoop, M*R^2.

The total kinetic energy is converted to potential energy when the hoop stops. Its potential energy is M*g*h, where h is the height of the hoop above the horizontal portion. If the hoop travels a distance L up the ramp, it climbs an amount = L*sinø. Now equating the potential energy at the top to kinetic energy at the bottom you get

.5*M*[2*π*R*w]^2 + .5*M*R^2*w^2 = M*g*L*sinø

You have all the data to solve for L

Answer 2

the linear velocity of the hoop is equal to the angular velocity multiplied by the radius of the hoop. that is 9 m/s. for rolling up the incline of 20 degrees, the retardation experienced by the hoop will be sine of 20 degrees into the accelaration due to gravity (9.8 m/s2). applying the basic formulae of constant acceleration we get the distance as 0.082m. that is along the incline.

Answer 3

Newton's 2d regulation gives you you the translational acceleration. Angular acceleration = translational acceleration / radius. The radius is the radius the door swings by, or merely its length (2.4 m). Viola. notice that i'm assuming the sections are arranged in a sq. (2 sections by 2 sections).