Rolling time

Assume line joining the centre of mass of the sphere and the centre of the track with horizontal be @

And the angular displacement of the sphere be #

For the surfaces are perfectly rough, no slipping occurs.

Hence, (R - r)@ = r# ... (1)

By energy conservation,

Loss in G.P.E. = Gain of K.E.

Mg(R - r)sin@ = 1/2 I(d#/dt)^2 + 1/2 Mv^2

where v is the linear speed of the sphere, v = (R - r)d@/dt

So, Mg(R - r)sin@ = 1/2 [2/5 Mr^2(d#/dt)^2 + M(R - r)^2(d@/dt)^2]

Mg(R - r)sin@ = M/2 [2/5 (R - r)^2(d@/dt)^2 + (R - r)^2(d@/dt)^2]

gsin@ = (R - r)/10 (d@/dt)^2

d@/dt = sqrt[10gsin@/(R - r)]

S (@ 0→pi/2) d@/sqrt(sin@) = sqrt(10/(R - r)) S (0→T)dt

Time, T = {sqrt[(R - r)/10]} S (@ 0→pi/2) d@/sqrt(sin@)

I leave the integral form of the solution since it seems analytically impossible to integrate the integral.

這裏可以幫到你
http://actioni.w7.c361.net/yahoo.com.hk/hk/auction/178987536

如果用Gamma function來表示的話,可以寫成:

(1/4) sqrt{(R-r)/(5pi)} Gamma^2(1/4) (s)

又或者

= k*sqrt{R-r} (s),其中 k=0.8291674...