Question

What is the minimum kinetic energy of the block at B, which is located at the top of the loop, so that the block can pass by this point without falling off from the track?

Answer 1

Answer:

${K.E = \dfrac{1}{2} mgR$

Explanation:

For loop-the-loop, to stay at the top of the loop the centripetal force must equal the gravitational force

$mg = \dfrac{mv^2}{R}$

solving for velocity we get:

$v = \sqrt{gR}$

Thus, the minimum kinetic energy required is

$K.E = \dfrac{1}{2} mv^2$

$\boxed{K.E = \dfrac{1}{2} mgR}$