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: Understanding the Concept:
The mass sits still at a fixed point \(P\) on the \(x\)-axis of the inertial frame \(S\), at distance \(r\) from the common origin, so it has zero velocity and zero acceleration in \(S\). Frame \(S'\) rotates relative to \(S\) with angular velocity \(\vec\Omega = \omega \hat z\). Because \(S'\) is non-inertial, an observer riding on \(S'\) must add pseudo-forces (centrifugal and Coriolis) to make Newton's second law work for this same mass.

Step 2: Key Formula or Approach:
1. Velocity seen in the rotating frame: \(\vec v_S = \vec v_{S'} + \vec\Omega \times \vec r\).
2. Centrifugal force: \(\vec F_{cen} = -m\vec\Omega\times(\vec\Omega\times\vec r)\), magnitude \(m\omega^2 r\), directed away from the axis.
3. Coriolis force: \(\vec F_{cor} = -2m\vec\Omega\times \vec v_{S'}\).

Step 3: Detailed Explanation.
Since the mass is at rest in \(S\), \(\vec v_S = 0\). Using the velocity relation above with \(\vec r = r\hat x\):
\[ 0 = \vec v_{S'} + \omega\hat z \times r\hat x = \vec v_{S'} + \omega r\hat y \]
\[ \Rightarrow \vec v_{S'} = -\omega r\hat y \]
This makes sense: since \(S'\) spins one way relative to \(S\), a point that is actually still in \(S\) appears in \(S'\) to trace a circle of radius \(r\) at angular speed \(\omega\), in the opposite rotational sense.
Now compute the centrifugal force:
\[ \vec F_{cen} = -m(\omega\hat z)\times\big[(\omega\hat z)\times(r\hat x)\big] = -m\omega^2(\hat z\times(\hat z\times r\hat x)) = -m\omega^2 r(\hat z\times \hat y) = m\omega^2 r\hat x \]
So \(F_{cen} = m\omega^2 r\), pointing radially outward.
Now the Coriolis force, using \(\vec v_{S'}=-\omega r\hat y\):
\[ \vec F_{cor} = -2m(\omega\hat z)\times(-\omega r\hat y) = 2m\omega^2 r(\hat z\times \hat y) = -2m\omega^2 r\hat x \]
So \(F_{cor} = 2m\omega^2 r\), pointing radially inward.
Comparing the two magnitudes:
\[ F_{cen} = m\omega^2 r, \qquad F_{cor} = 2m\omega^2 r \quad \Rightarrow \quad F_{cen} = \frac{F_{cor}}{2} \]

Step 4: Why the other options are wrong.
Both forces are nonzero because \(P\) is displaced from the rotation axis and does have a nonzero apparent velocity in \(S'\), so option (A) fails. Options (C) and (D) get the ratio backwards or equal; the actual ratio is \(1:2\), not \(2:1\) or \(1:1\).

Step 5: Final Answer:
Both pseudo-forces are nonzero, and the centrifugal force is exactly half the Coriolis force.\[ \boxed{\text{(B)}} \]