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# Loop-The-Loop

Let's discuss an interesting problem - Loop-The-Loop, useful for Physics Olympiad.

The Problem:

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
$$F=\sqrt{(8mg)^2+(mg)^2}=\sqrt{65}mg$$

# Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy