Question

Two frames \(S\) (solid lines) and \(S'\) (dashed lines) with common origin are shown in the figure below. Frame \(S\) is inertial while \(S'\) is rotating about the common \(z\)-axis. There is a point mass fixed at \(P\) on the \(x\)-axis of the \(S\) frame. The magnitude of the centrifugal force and the Coriolis force experienced by the mass in the \(S'\) frame is \(F_{cen}\) and \(F_{cor}\), respectively. Which of the following options is correct for these forces?

• A. \(F_{cen} = 0\) and \(F_{cor} = 0\)
• B. \(F_{cen} \neq 0\) and \(F_{cor} \neq 0\) and \(F_{cen} = \dfrac{F_{cor}}{2}\)
• C. \(F_{cen} \neq 0\) and \(F_{cor} \neq 0\) and \(F_{cen} = 2F_{cor}\)
• D. \(F_{cen} \neq 0\) and \(F_{cor} \neq 0\) and \(F_{cen} = F_{cor}\)

Hint:

A mass fixed in the inertial frame \(S\) appears in the rotating frame \(S'\) to move on a circle of radius \(r\) at speed \(\omega r\); use \(F_{cen}=m\omega^2r\) and \(F_{cor}=2m\Omega v' = 2m\omega^2r\) to get the ratio \(1:2\).

Answer 1

The Correct Option is B

Step 1: Picture the motion directly as seen from the rotating frame.
Instead of starting from the velocity-addition formula, think about what an observer sitting on $S'$ actually sees. The mass at $P$ never moves in the inertial frame $S$, but since $S'$ itself spins relative to $S$ at rate $\omega$, the $S'$-observer sees $P$ trace out a circle of radius $r$ (the fixed distance of $P$ from the common origin) at angular speed $\omega$, going around in the sense opposite to how $S'$ spins relative to $S$. This is a purely kinematic fact, true even before any forces are considered.

Step 2: Get the speed and true (kinematic) acceleration of this apparent circular path.
For circular motion of radius $r$ at angular speed $\omega$, the speed is $u = \omega r$ and the acceleration needed to keep tracing that circle points toward the center with magnitude $a' = u^2/r = \omega^2 r$. This is simply the standard centripetal acceleration formula for any object moving on a circle, applied to the apparent path of $P$ as tracked using $S'$'s own axes.

Step 3: Match this kinematic acceleration to the pseudo-forces.
Since $P$ feels zero real force (it just sits still in the inertial frame $S$), Newton's second law written in the non-inertial frame $S'$ must produce this same net inward acceleration purely from pseudo-forces: $m a'_{\text{net}} = F_{cor} - F_{cen}$, where $F_{cor}$ points inward (since it opposes the apparent circular drift) and $F_{cen}$ points outward. Using the standard expressions $F_{cen} = m\omega^2 r$ and $F_{cor} = 2m\omega^2 r$ (from $F_{cor} = 2m\omega u = 2m\omega(\omega r)$, since the Coriolis force magnitude is $2m\Omega v'$ for velocity perpendicular to the rotation axis):
\[ F_{cor} - F_{cen} = 2m\omega^2 r - m\omega^2 r = m\omega^2 r = m a'_{\text{net}} \]
which exactly matches $m a' = m\omega^2 r$ found by pure kinematics in Step 2. This cross-check confirms both magnitudes are consistent with each other.

Step 4: Read off the ratio.
From $F_{cen} = m\omega^2 r$ and $F_{cor} = 2m\omega^2 r$, both are clearly nonzero, and dividing gives $F_{cen}/F_{cor} = 1/2$, i.e. $F_{cen} = F_{cor}/2$.

Final Answer:
Both pseudo-forces are present and the centrifugal force is exactly half the Coriolis force.\[ \boxed{\text{(B)}} \]