Consider a bob whose upper and is fixed with length of 2 m is given with a horizontal push through angular displacement of `60^@` . What is the angular velocity of the bob ? If it is of 2kg . [Take ,g`=10ms^(-2)]`

Options

A. `sqrt10rads^(-1)`
B. `sqrt20rads^(-1)`
C. `sqrt8rads^(-1)`
D. None of these

Answer

A

Explanation

Given , mass ,= 2kg
angular displacement ,`theta=60^@`
length,`l=2m and g=10ms^(-2)`

From FBD, we have `Tcostheta=mg" "(i)`
`Tsintheta=mgomega^(2)" "(ii)`
From Eqs.(i) and (ii) ,We get
`omega=sqrt((g tantheta)/(r))`
Also " " `r=lsintheta`
`therefore" "omega=sqrt((gsintheta)/(costhetaxxlsintheta``omega=sqrt((g)/(lcostheta))=sqrt((10)/(2xxcos60^@))=sqrt(10)/(1)``rArr " " omega=sqrt10rads^(-1)`