EEEK conical pendulum ... period of motion. (1 Viewer)

[haboozin]

A body of mass 0.5kg is suspended from a fixed point O by means of a light rod of leng 1m. the mass is rotated in a horizontal circle at a constant speed and the rod makes an angle of 30* with the downward direction of the virtical ... g = 9.8ms^-2


iv) find the period of the motion.

?????

 

[HSCExamMarker]

umm..

sin 30 = r/1
r = 0.5
m = 0.5

Tsin30 = mrw^2 (1)
= 0.5*0.5 *w^2
Tcos30 = 9.8 * 0.5 (2)

for (1)
T = w^2/2

for (2)
T = 9.8/root3

then u can find w (angular velocity) rite? by solving T

Then just Period = 2pi/w

If i am wrong..plz correct me :d

 

[Bellow]

i dont have a diagram here but if u do u'd prolly get it,

Tcos30 = mg
Tsin 30 = mv^2/r (since v=rw, where w = omega)
= mrw^2

divide these equations to eliminate T,
therefore,
tan 30 = rw^2/g
now, sin 30 = r/1 = 1/2
so, tan 30 = 1/2 x w^2/9.8
1/sqr root 3 = 1/2 x w^2/9.8
w^2 = 19.6/sqrroot 3
w = 3.3639.....
now, P = 2(pi)/w
therefore P = 2(pi)/3.3639....
= 1.8678....
= 1.9

is this correct? my conical pendulum still weak

 

[haboozin]

oh 2pi/w

that gay thing oopsie..