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(C) The plank is resting on the Earth's surface,which is rotating about its axis with angular velocity $\omega$. The plank undergoes circular motion w

Vedclass · Accepted Answer

$\frac{m r_e \omega^2}{2}$

Vedclass · Answer

(C) The plank is resting on the Earth's surface,which is rotating about its axis with angular velocity $\omega$. The plank undergoes circular motion with radius $r = r_e \cos 45^{\circ}$.The centripetal force required for this circular motion is $F_c = m \omega^2 r = m \omega^2 (r_e \cos 45^{\circ})$.This centripetal force acts towards the axis of rotation. The horizontal ground at latitude $45^{\circ}$ is inclined at an angle of $45^{\circ}$ with the axis of rotation. The component of the centripetal force acting parallel to the ground provides the necessary frictional force to keep the plank at rest relative to the ground.$f = F_c \sin 45^{\circ} = (m \omega^2 r_e \cos 45^{\circ}) \sin 45^{\circ}$Since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have:$f = m \omega^2 r_e \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) = \frac{m r_e \omega^2}{2}$

$A$ plank is resting on a horizontal ground in the northern hemisphere of the earth at a $45^{\circ}$ latitude. Let the angular speed of the earth be $\omega$ and its radius $r_e$. The magnitude of the frictional force on the plank will be

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