# A thin ring, a disk and an annular cylinder, of same mass M, are released from a point 3.6 m from the ground up an inclined plane of 30o inclination. The ring and the disk have the same radius R. For the annular cylinder, the outer radius is R and the inner radius is R/2. Times taken by the ring, disk and annular cylinder, respectively, to reach the ground are in the ratio,1.√2 : √1.5 : √1.6 2. √1.4 : √1.5 : √2 3.√2 : √1.5 : √1.5 4.√1.4 : √1.5 : √1.6

Case of ring I = MR2

Mgy = ½Mv2 + ½Iω2

=> Mgy = ½Mv2 + ½(MR22

=> Mgy = ½Mv2 + ½(MR2 )(v/R)2

=> Mgy =  ½Mv2 + ½Mv2

=> v2 = gy

=> v2  = gy

=> v = √(gy)

Time taken,

Case of disk I = ½MR2

Mgy = ½Mv2 + ½Iω2

=> Mgy = ½Mv2 + ½(½MR22

=> Mgy = ½Mv2 + 1/4MR2(v/R)2

=> Mgy =  3/4Mv2

=> v2 = 4/3gy

=> v  = √(4/3)gy

Case of annular cylinder I = ½(r12 + r22)

r1 = R

r2 = ½R

I = 5/8R2

Mgy = ½Mv2 + ½Iω2

=> Mgy = ½Mv2 + ½(5/8R22

=> Mgy = ½Mv2 + ½(5/8MR2 )(v/R)2

=> Mgy =  ½Mv2 + 5/16Mv2

=> 13/16 v2 = gy

=> v2  = (16/13)gy

=> v =√{(16/13)gy}

Taking the ratios of time t in three cases

We get,

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