A flywheel of mass 1 kg and radius vector `(2hati _ hatj + 2 hatk)` m is at rest. When a force `(3hati + 2 hatj - 4 hatk)` N acts on it tangentially, it can rotate freely. Then , its angular velocity after `4.5s` is

Options

A. `2/9 sqrt( 261)" rad " s^(-1)`
B. `3/2 sqrt( 261)" rad " s^(-1)`
C. ` sqrt( 261)" rad " s^(-1)`
D. `5/9 sqrt( 261)" rad " s^(-1)`

Answer

C

Explanation

Given, mass of flywheel, `M = 1 ` kg,
radius vectors `R = (2 hati + hatj + 2hatk)` m
`F = (3 hati + 2hatj + 4hatk)` N and time `t = 4.5s`
Magnitude of radius
`(R) = sqrt((2)^(2) + (1)^(2) + 2^(2)) = sqrt(9) = 3 ` m
Similarly, ` F = sqrt(29) N`
Torque on the flywheel,
`tau = I alpha = F. R = (MR^(2))/2 alpha `
`alpha = (2F)/(MR) = (2sqrt(29))/(Ixx3) = 3/2 sqrt(29)`
Now, the angular velocity,
`omega = omega_(0) + alpha t rArr omega =0 + 2/3 sqrt(29) xx 4.5 (because omega _(0) = 0)`
`rArr omega = sqrt(261 )" rad " s^(-1)`
Hence, the correct option is (c).