Rotational Mechanics..2nd Question..!!

Please help me with a detailed explanation..!

using torque about bottommost point ,
T = I * a/r .... (assuming pure rolling)
now T= mgrsinθ
I = c . mr² ( c depends upon the object)

replacing, we have,
mgrsinθ = c. mr² . a /r

g sinθ = c.a
or, a = gsinθ /c
hence the object having greater value of c about bottom most point , will take more time and vice versa.

values of c = 1.ring - 3/2
2. cylinder - 3/2
3. hollow sph. - 5/3
4. solid sph. - 7/5

hence ring and cylinder both will reach the bottom at last.

Thank You Aditya it was a wonderful explaination..!!...and you are ofcourse right in this case..!!:P

no, if I =Mk² , then k =radius of gyration
here i have taken I =c. Mr² = M . (√c.r)²
hence radius of gyration ,k in this case is √c.r

Ohho... Miss kar gaya. Bas coefficient ko ' c' likha hai.. toh thik hai..!!

lol!
btw kaushal answer will be hollow sphere , c= 1.66 which is greater than ring ,c =1.5