Question

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The marble is initially at rest at the top of the ramp (point A), a height h above the end of the ramp. When it reaches the bottom of the ramp, it has some speed v (which we need to know to predict the landing point on the floor). So, the initial potential energy of the marble at point A has been transformed into kinetic energy. Its important to recognize that its kinetic energy is in two forms: .o Table Top translational kinetic energy, given by Kmv* rotational kinetic energy (because the marble is rolling, not just sliding K- -ya, where l is the moment of inertia of the marble (a solid sphere), and o is its rotational speed By knowing the equation for the moment of inertia of a sphere we can derive an equation for the exit velocity of the marble in terms of the height h. Let the moment of inertia of the sphere be /-x.mR2 where x is some number that depends on whether the sphere is hollow or solid. Prelab Assignment: Use conservation of energy to derive an algebraic equation for the speed of the marble Your final answer should start with "v"The other side may include only "g" and things you can measure with your equipment. It may include the mass of the marble, the radius of the marble, the height and angle of the ramp. Nothing else. (You will not need all of these variables), Make your final expression as simple as possible. Try to get your answer in t mms of only ..g ..h", and "x" (from lex. mR2).

Answer 1

Using Conservation of the energy mgh = mV^2/2 + Iw^2/2 -(1) we know that I = XMR^2 w = V/R Put these values in equation (1) mgh = m v^2/2 + 1/2 [X MR^2 middot V^2/R^2] gh = V^2/2 + XV^/2 gh = (1 + X) v^2/2 V^2 = 2 gH/ 1 + X V = squareroot 2gh/ 1 + X