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(MR2 )ω2
=> 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(½MR2)ω2
=> 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(5/8R2)ω2
=> 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,
