Question

A mass m slides down a frictionless ramp and approaches a frictionless loop with radius R. There is a section of the track with length 2R that has a kinetic friction coefficient of 0.5. From what height h must the mass be released to stay on the track?

Answer 1

Answer:

Minimum velocity required to complete verticle circle of radius R at lowest point is

v=√(5gR)

or,v²=5gR

this is the velocity which is attented by crossing segment having  kinetic friction coefficient of 0.5 of length 2R,

while crossing 2R its velocity gets reduced

let,velocity just before crossing 2R be u then

ma=μN (N=normal reaction)

ma=μmg

or,a=μg

now we have

v²=u²-2a(2R)

u=√(v²+4aR)

 =√(5gR+4μgR)

  =√(5gR+2gR)

  =√(7gR)

from conservation of energy

mgh=(1/2)m*u²

gh=(1/2)*(7gR)

h=(7/2)R

so From height h=(7/2)R must the mass be released to stay on the track