Cheenta
How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

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
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com