This applet shows the same situation, but it also shows, through bar graphs that

Question

This applet shows the same situation, but it also shows, through bar graphs that change with time, the way that the enemy is transformed as the box and the disk go down the inclined plane. Assume that the box and disk each have m, the top of the incline is at height h, and the angle between the incline and the ground is theta (i.e., the incline is at an angle theta above the horizontal). Also, let the radius of the disk be R. Part B How much sooner does the box reach the bottom of the incline than the disk? Express your answer in terms of some or all of the variables m, h, theta, and R, as well as the acceleration due to gravity g

Explanation / Answer

The block slides down the incline without rotating. Its acceleration is : aB=g* sin? The disk rolls down the incline without sliding. Its acceleration is : aD=g* sin? / [1+(I / MR^2)] I for cylinder = 0.5 MR^2....so aD = 2 * g * sin? / 3 time taken to reach bottom = sqrt [ 2* L / a] = sqrt [ 2 * h / a * sin?] so, time taken by block = [2h / g * sin^2 ?]^0.5 time taken by disk = [3h / g * sin^2 ?]^0.5 So the block will reach : 0.3178 *[h / g * sin^2 ?]^0.5

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