Consider a wheel rotating around a fixed axis. If the rotation angle ` theta` varies with time as `theta = at^2` then the total acceleration of a point A on the rim of the wheel is (v being the tangential velocity)

Options

A. `v/t sqrt(1 + 4a^2t^4)`
B. `v/t `
C. `v/t (1 + 4a^2 t^4)`
D. `sqrt(1 + 4a^2t^4)`

Answer

A

Explanation

Given , ` theta = at^2`
so , angular velocity , `omega = (d theta)/(dt) = 2 at`
Linear tangential velocity
` v = omega r = 2 at r (v)/(t) = 2ar`
so, tangential linear acceleration of particles is
`a_1 = (dv)/(dt) = 2ar`
and angular acceleration of particles is
`alpha = (domega)/(dt) = 2a`
also , normal or radial acceleration of particle is
` a_n = (v^2)/(r) = (4a^2 t^2 r^2)/(r) = 4a^2 t^2 r`
total acceleration of particle is
`a_"total" = sqrt(a_t^2 + a_n^2) = sqrt( 4a^2 r^2 + 16a^4 t^4 r^2)`
` = 2ar sqrt(1 + 4a^2 t^4)`
` v/t sqrt(1 + 4a^2 t^4)`