A particle of mass 1 kg, tied to a 1.2 m
long string is whirled to perform vertical
circular motion, under gravity. Minimum
speed of a particle is 5 m/s. Consider
following statements.
P) Maximum speed must be 5 5 m/s.
Q) Difference between maximum and
minimum tensions along the string is 60 N.
Select correct option.
(A) Only the statement P is correct.
(B) Only the statement Q is correct.
(C) Both the statements are correct.
(D) Both the statements are incorrect.
Answers
it's interesting question....
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=1×5×5×10/12 -1×10
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+2×10×2×1.2
V^2=25+48
V=√73
let the tension be t2 at tension at the bottom most point,
t2=mg+mv^2/R
t2=1×10+1×73/1.2
t2=10+6.8
t2=7.8
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 √73,=8.5
so option p is incorrect...
(hence option A is right)
xD skipping rotational dynamics is the best exam strategy baaki toh..khair chhodiye xD milte hai agli baar,tab tak ke liye keep missing me.
