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Question: A rod of length L is rotated in horizontal plane with constant angular velocity ω. A mass m is suspe...

A rod of length L is rotated in horizontal plane with constant angular velocity ω. A mass m is suspended by a light string of length L from the other end of the rod. If the angle made by vertical with the string is θthen angular speed, ω =

A

[gsin⁡θL(1+tan⁡θ)]1/2\left[ \frac { \mathrm { g } \sin \theta } { \mathrm { L } ( 1 + \tan \theta ) } \right] ^ { 1 / 2 }

B

[L(1+tan⁡θ)gtan⁡θ]1/2\left[ \frac { \mathrm { L } ( 1 + \tan \theta ) } { \mathrm { g } \tan \theta } \right] ^ { 1 / 2 }

C

[gtan⁡θL+sin⁡θ]1/2\left[ \frac { \mathrm { g } \tan \theta } { \mathrm { L } + \sin \theta } \right] ^ { 1 / 2 }

D

[gtan⁡θL(1+sin⁡θ)]1/2\left[ \frac { g \tan \theta } { \mathrm { L } ( 1 + \sin \theta ) } \right] ^ { 1 / 2 }

Answer

[gtan⁡θL(1+sin⁡θ)]1/2\left[ \frac { g \tan \theta } { \mathrm { L } ( 1 + \sin \theta ) } \right] ^ { 1 / 2 }

Explanation

Solution

Radius of horizontal circle of ball = (L + L sinθ)

∴ C.P. Acceleration = (L + L sin θ)ω2 ( a = rω2)

Here mg = T cos θ …….. (i)

and mω2(L + L sin θ) = T sin θ ……… (ii)

Dividing (ii) by (i)

tan θ = ω2( L+Lsin⁡θ)g\frac { \omega ^ { 2 } ( \mathrm {~L} + \mathrm { L } \sin \theta ) } { \mathrm { g } }

or ω2 = gtan⁡θL(1+sin⁡θ)\frac { g \tan \theta } { \mathrm { L } ( 1 + \sin \theta ) }