- #1
jiboom
- 91
- 0
one end of a light inelastic string is attached to a point A vertically above a point O on a smooth horizontal plane and at a height h above it. The string carries a particle P of mass m at its other end. WHEN JUST TAUT the string is inclined to the vertical at an angle S
if P moves in a horizontal circle,centre O,with speed v,show that
v^2<= hg tan^2 s
im reading this as the particle is on the smooth plane,as the circle centre is O,so my equations of motion are,with T=tension,r=rad of circle,R= reaction force
R+Tcos S-mg=0
Tsin S=mv^2/r
i know r=htan S
i know at speed V string is just taut,but putting T=0 makes no sense, so how to proceed?if i put T>0 i still need to get rid of the reaction force.
regarding the just taut part. on another question i had 2 strings attached to a particle at one end, and the other ends of the strings were attached at points in a vertical line. the paricle moved in circle at a speed which just made strings taut. to do the question i had to let the tension in the bottom string = 0. so what does just taut mean?
if P moves in a horizontal circle,centre O,with speed v,show that
v^2<= hg tan^2 s
im reading this as the particle is on the smooth plane,as the circle centre is O,so my equations of motion are,with T=tension,r=rad of circle,R= reaction force
R+Tcos S-mg=0
Tsin S=mv^2/r
i know r=htan S
i know at speed V string is just taut,but putting T=0 makes no sense, so how to proceed?if i put T>0 i still need to get rid of the reaction force.
regarding the just taut part. on another question i had 2 strings attached to a particle at one end, and the other ends of the strings were attached at points in a vertical line. the paricle moved in circle at a speed which just made strings taut. to do the question i had to let the tension in the bottom string = 0. so what does just taut mean?