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# Loop-The-Loop

A small block of mass $$m$$ slides along the frictionless loop-the-loop track . If it starts at $$A$$ at height $$h=5R$$ from the bottom of the track then what will be the resulting force acting on the track at $$B$$?

Discussion:

Let the velocity at $$B$$ be $$v$$.
Now,
kinetic energy $$\frac{1}{2}mv^2$$=potential energy lost $$mg(5R-R)$$
$$\frac{mv^2}{R}=8mg$$
This is the centrigular force acting horizontally,
The weight acts vertically down.
Hence, the resultant force
$$F=\sqrt{(8mg)^2+(mg)^2}=\sqrt{65}mg$$

September 3, 2017