IIT JEE , AIEEE Forum - Physics , Mechanics

gabriel

A disc of radius R has a light pole fixed

to the disc at the circumference which in turn has a pendulum of lenght R attached to its other end.The disc is rotated with a constant angular velocity w.the string is making an angle 30with teh rod .The angular velocity w of the disc is

a) g ^{[ 2]}

3   ^1/2
     -------
        R
b) g^{[2 ]}

3   ^1/2
        -------
        2R

c)       g    ^1/2
        -------
         R ^{[2 ]}

3

d) 2g ^1/2
---------
3R^[2]

3

rajat

i presume that the disc is rotating about its axis .

answer is : d

truly

the disk rotates about its axis....
the horizontal distance between the center of disk and bob of pendulum = R+Rsin30 = 3R/2

F=force on it due to rotation = mw²(3R/2)

balancing torque about the point at which it is suspended....T1 = T2
torque due to weight = T1 = mg Rsin30 = mgR/2
torque due to F=> T2 = F Rcos30 =(FR

3)/2

T1 = T2
(FR

3)/2 = mgR/2
=> F = mg/

3
=>3mw²R/2 = mg/

3
==>3w²R/2 = g/

3
=> w² = 2g / 3R

3
=> w =

[2g / 3R

3]

hence option d is correct

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