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A small, solid sphere of mass 0.2 kg and radius 40 cm rolls without slipping along the track consisting of slope and loop-the-loop with radius 1.5 m at the end of the slope. It starts from rest near the top of the track at a height h, where h is large compared to 40 cm. What is the minimum value of h (in terms of the radius of the loop R) such that the sphere completes the loop?

Respuesta :

Answer:

Explanation:

Radius of loop = R

minimum value of velocity at the base of the loop so that it can complete the loop

= √ 5gR .

If h be the height from where the fall gives this value of final velocity

v = √ 2gh

So ,

√ 2gh  = √ 5gR .

2gh = 5gR

h = 5 /2 R

Putting the value of R = 1.5 m ( given in the problem )

h = 2.5 x 1.5 = 3.75 m .

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