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The distance moved by a particle in tim...

The distance moved by a particle in time from centre of ring under the influence of its gravity is given by `x=a sin omegat` where a and `omega` are constants. If `omega` is found to depend on the radius of the ring (r), its mass (m) and universal gravitation constant (G), find using dimensional analysis an expression for `omega` in terms of r, m and G.

Text Solution

Verified by Experts

The correct Answer is:
`sqrt((GM)/(r^(3)))`
`omegapropr^(a)m^(b)G^(c)`
`omega =k[L]^(a)[M]b[M^(-1)L^(3)T^(-2)]^(c) " " [krarr"Proporsionality constant"]`
`T^(-1)=kL^(a+3c)M^(b-c)T^(-2c)`
`c=(1)/(2), b=(1)/(2), a=-(3)/(2)`
`omega =ksqrt((GM)/(r^(3))`
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