Question

A girl is rotating a hoop with diameter d = 90 cm round waist. The waist size of the girl is l = 75 cm. How many rotations per minute shall do the hoop so that not to fall down? The friction coefficient equals 0.5, the gravity acceleration g =10 m/s2.

Answer 1

The hoop is effected by the gravity force, the normal reaction and friction forces at the point where the hoop is currently touching the girl’s waist:

$mg=F_{friction} (1)$

$mg=kmω^2(D-d) (2)$

In our case, we have g =10 m/s², k=0.5, D=0.45 m, d=0.375 m

Using (2) we get

$ω=\sqrt{\frac{g}{k(D-d)}} (3)$

Using (3) we get

ω =16.33 rad/s (4)

The rotations per second is equal to

$n=\frac {ω}{2π} (5)$

We put (4) in (5)

n=2.6 rotations per second

1 minute has 60 second.

In this way, the number of rotations per minute is equal to

2.6×60=156

Answer

156