1Please help me with a detailed explanation..!
262using 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.
1
kaushalop07
·2011-06-25 02:31:36
Thank You Aditya it was a wonderful explaination..!!...and you are ofcourse right in this case..!!:P
262
Aditya Bhutra
·2011-06-25 06:50:48
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
71Ohho... Miss kar gaya. Bas coefficient ko ' c' likha hai.. toh thik hai..!!
262
Aditya Bhutra
·2011-06-25 21:33:39
lol!
btw kaushal answer will be hollow sphere , c= 1.66 which is greater than ring ,c =1.5
