Non-inertial Frames Questions and Answers

Answer: B

Both forces are non-zero, with \(F_{cen} = F_{cor}/2\).

In the rotating frame S' the mass, fixed in S at distance r from the axis, appears to move in a circle at speed \(v' = \omega r\), opposite to the frame's rotation. The fictitious forces are:

\[ \begin{aligned} F_{cen} &= m\omega^{2}r\ \text{(outward)} \\ F_{cor} &= 2m\omega v' = 2m\omega^{2}r\ \text{(inward)} \end{aligned} \]

Their sum is \(m\omega^{2}r\) inward, exactly the centripetal force needed for the circular motion seen in S', with no real force acting. The forces are not zero, because S' is rotating, and the Coriolis force is twice, not half or equal to, the centrifugal force, which rules out the remaining options.

Answer: B

The period is inversely proportional to sin \(\phi\).

The plane of a Foucault pendulum turns at the angular rate \(\Omega\sin\phi\), where \(\Omega\) is the earth's rotation rate, because only the vertical component of the earth's rotation turns the plane. The time for one complete turn is:

\[ \begin{aligned} T &= \frac{2\pi}{\Omega\sin\phi} \\ &= \frac{24\ \text{h}}{\sin\phi} \end{aligned} \]

At the poles (\(\phi = 90\)°) the plane turns once a day, and at the equator it does not turn at all. The period is not proportional to the latitude itself, nor to \(\cos\phi\).

Answer: C

The derivative in the rotating frame is \(\frac{d'\bar{A}}{dt} = -\bar{\omega} \times \bar{A} + \frac{d\bar{A}}{dt}\).

The standard operator relation between a fixed frame S and a frame S′ rotating with angular velocity \(\bar{\omega}\) is

\[ \begin{aligned} \left(\frac{d\bar{A}}{dt}\right)_{S} &= \left(\frac{d\bar{A}}{dt}\right)_{S'} \\ &\quad + \bar{\omega} \times \bar{A} \end{aligned} \]

Solving for the rotating-frame derivative gives option C. Option B has the wrong sign of the rotation term, option A drops the fixed-frame rate of change, and option D would hold only if \(\omega = 0\). A vector fixed in S, for example, appears in S′ to rotate backwards at \(-\bar{\omega} \times \bar{A}\).