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?
Answers
Answer:
Frictional force = f = mg
Centripetal Force = Contact force = N = mv² / r
μ = f / N
= mg / (mv^2 / r)
= rg / v²
= (5 × 10) / (5√5)²
= 50 / 125
= 0.4
Minimum coefficient of friction between the tyres and wall of the well must be 0.4.. HOPE IT WOULD HELP YOU MAKE ME BRAINIEST....
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
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