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Question

A uniform disc of mass m and radius R is connected with two light springs 1 and 2 . The springs are connected at the highest point M and the center of mass (CM) 'N' of the disc. The other ends of the springs are rigidly attached with vertical walls. If we shift the CM in horizontal direction by a small distance, the disc oscillates simple harmonically. Assuming a perfect rolling of the disc on the horizontal surcface, the angular frequency of oscillation is (answer upto two decimal places) ( Take k1 = 10 N/m , k2 = 20 N/m, m=10 kg )

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Solution

Let us displace the CM by a small distance x towards right. As a result the body rolls on the horizontal surface. Hence the spring 1 is elongated by x1=2Rθ and spring 2 is compressed by x2=Rθ, where θ= angle of rotation of the rolling body.


The spring forces F1 and F2 produce anticlockwise torques F1(2R) and F2(R) respectively about P . Summing up the spring torques about P , we have τnet =F1(2R)+F2(R) Where F1=k1x1=k1(2Rθ) and F2=k2x2=k2Rθ
This gives
τnet=(4R2k1+R2k2)θ=(4k1+k2)R2θThis torque produces anticlockwise acceleration α for clockwise displacement θ, which can be given as
α=τnetIP=(4k1+k2)R2θIP
Substituting IP=mR2+IC=mR2+mR22=3mR22
We have α=2(4k1+k2)3mθ
Comparing the above equation with a=ω2θ
We have ω=2(4k1+k2)3m=2(4×10+20)3×10=2

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