a block of mass m slides down on a frictionless loop-the-loop track of radius R

Question

a block of mass m slides down on a frictionless loop-the-loop track of radius R as shown. (1) When the block is released from rest at height h above the ground, it passes point P which makes an angle theta with the vertical (as shown). Find the velocity at point P as a function of h,theta, and R. (2) Find the minimum height h such that the block would be able to complete the loop. (3) When it is released with that minimum height, it completes the loop, and shoots out into the air at an angle of 30 degrees from horizontal at the bottom(as show in the figure). Find the maximum height (in the air) that the block would go up to. (4) Find the distance D where the block lands on the ground.

Explanation / Answer

1) Here,    mgh = 1/2mv2 + mg * ( R - Rcos(theta))

=>          gh    - g * ( R - Rcos(theta)) = 1/2v2

=>   v = sqrt[2g * (h - R + Rcos(theta))]

=>    velocity at point P =    sqrt[2g * (h - R + Rcos(theta))]

2)     Here, mgh    = mg * 2R + mgR/2

     => h =   (5/2)R

=>    the minimum height h   =    (5/2)R

3) Maximum height = Rg * 0.25/(2 * g)  

                                 =   0.125 * R

                                 = 0.125R

4)    distance D where the block lands on the ground =     Rg * 0.866/g

                                                                                    =   0.866R

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