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(B) In the rotating frame of the table,the ball experiences a centrifugal force $F_c = m \omega^2 x$,where $x$ is the distance from the axis of rotati

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) In the rotating frame of the table,the ball experiences a centrifugal force $F_c = m \omega^2 x$,where $x$ is the distance from the axis of rotation.Since the groove is smooth,the acceleration of the ball along the groove is $a = \frac{F_c}{m} = \omega^2 x$.We know that $a = v \frac{dv}{dx}$,where $v$ is the radial velocity.So,$v \frac{dv}{dx} = \omega^2 x$.Integrating both sides with respect to $x$ from the initial position $x_i = 1 	ext{ m}$ to the final position $x_f = 3 	ext{ m}$:$\int_0^v v \, dv = \int_1^3 \omega^2 x \, dx$$\frac{v^2}{2} = \omega^2 \left[ \frac{x^2}{2} ight]_1^3$$\frac{v^2}{2} = \frac{\omega^2}{2} (3^2 - 1^2)$$v^2 = \omega^2 (9 - 1) = 8 \omega^2$$v = \sqrt{8} \omega = 2 \sqrt{2} \omega 	ext{ m/s}$.Comparing this with $x \sqrt{2} \omega$,we get $x = 2$.

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