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You are testing a new roller coaster ride in which a car of mass m moves around a vertical circle of radius R. In one test, the car starts at the bottom of the circle (point A) with initial kinetic energy Ki. When the car reaches the top of the circle (point B), its kinetic energy is Ki/4, and its gravitational potential energy has increased by Ki/2. What was the speed of the car at point A, in terms of g and R?

Respuesta :

The initial velocity is 2.3 [tex]\sqrt{gR}[/tex]

Explanation:

Suppose the speed of car at point A is u .

Its kinetic energy K = [tex]\frac{1}{2}[/tex] mu²

At point B , its kinetic energy = [tex]\frac{1}{4}[/tex] K

and its gravitational potential energy = [tex]\frac{1}{2}[/tex] K = m g h = 2 m g R

here height h = 2 R ( radius of circle )

From the principle of conservation of energy

[tex]\frac{1}{2}[/tex] m u² =  [tex]\frac{1}{8}[/tex] m u² + 2 m g R

[tex]\frac{3}{8}[/tex] m u² = 2 m g R

Therefore u = [tex]\sqrt{\frac{16gR}{3} }[/tex] =  2.3 [tex]\sqrt{gR}[/tex]

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