Question

During a stunt, a cyclist (considered to

be a particle) is undertaking horizontal

circles inside a cylindrical well of

radius 6.05 m. If the necessary friction

coefficient is 0.5, how much minimum

speed should the stunt artist maintain?

Mass of the artist is 50 kg. If she/he

increases the speed by 20%, how much

will the force of friction be?​

Answer 1

Answer:

Vmin  = 11 m/s

f  = 500 N

Explanation:

Given data :-  

cylindrical well of radius r = 6.05 m

Coefficient of friction is μ = 0.5

Mass of the artist is m = 50 kg

What we have to find out:-

1)how much minimum speed should the stunt artist maintain Vmin =?

2) How much will the force of friction be If she/he increases the speed by 20%, f =?

Solution:-

The minimum velocity to maintain motion is given by

$v =\sqrt{(rg/u)}$

Vmin  = √( 6.05* 10/0.5) .............g = 10m/s2

   = √( 121)

 Vmin  = 11 m/s

In well of death, the force of friction id depend on weight, you can reffer from fig in textbook, hence we get,

      f = fs = mg = 50*10

 f  = 500 N