a particle of mass 1kg ,tied to a1.2m long string is whirled to perform vertical circular motion, under gravity.minimum speed of a particle is 5m/s. consider following statement p)maximum speed must be 5 under root 5 m/s Q)difference between maximum and maximum and minimum tension along the string is 60N select correct option
Answers
Answer:
minimum velocity will be at the maximum height...
Let the tension be t1 at the highest postion...
then,
clealry we can observe that
mg force will be in the downward direction...and at highest point, centripetal force will be in opposite direction....
so their, resultant,will be the tension...of the
string
t1=Mv^2/R -Mg
t1=1x5x5x10/12 -1x10
t1=250/12-10
t1=130/12=1.8(approx)
since the acceleration,will be constant
g,everywhere,so,we can take out the maximum velocity,which will be at downward position...
V^2=u^2+2x10x2x1.2
V^2=25+48
V=173
let the tension be t2 at tension at the bottom most point,
t2=mg+mv^2/Rt2=1x10+1x73/1.2
t2=10+6.8
t2=7.8
Vo) LTE+ LTE1
t2-t1=7.8-1.8
t2-t1=60
from this we get that, option Q is exactly right....
also,we take out maximum velocity as v73,=8.5
so option p is incorrect...