a small bead with mass of M is attached to aa very light string hung from a celiling.?

Question

The string can be torn by a force exceeding the value of 10 Mg where g is acceleration due to gravity. The bead on the string may ocsillate harmonically with a period T0. a student decides to perform another experiment. he takes the attched nead aside from its equilibrium and pushes it in such a way that the bead performs a full revolution in a horizontal plane. What is the time of one revolution of the bead that can be calculated with the data, given above? is this time a maximum or a minimum value of all possible?

Answer 1

The period of oscillation T is given by

T=2 pi sqrt(L/g)
L - length of the string
g - acceleration due to gravity(as you have put it)
pi =3.14159...

L=R=gT0^2/(2 pi)^2
R - radius of revolution



The max force should not exceed 10mg
F=ma'
F - centripetal force
a' - a + g (max acceleration at the bottom of the trajectory.
a - centripetal acceleration
m - mass in question

We can compute both the min time and the max time

For min time a=>g since at the very top we want the string to be just tight enough.

For max time;
since 10g will tear the string a max <= 9g

since we know a we can compute the speed

V=sqrt(a R)

and t=S/V

S= 2pi R

we have
t=2 pi R /sqrt( a R)
just to make a bit more eye pleasing
t=2 pi /sqrt( R/a )

t(min)=2 pi / sqrt(R/g)



t(max)=2 pi / sqrt(R/9g)


Just remember R=gT0^2 / (2 pi)^2