Non-inertial Frames Questions and Answers

Q: 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?

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.

Q: The period of oscillation of the plane of Foucault's pendulum is :(\(\phi\) is the latitude)

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 \(\cos\phi\).

Q: A frame of reference S′ is rotating with constant angular velocity \(\omega\) with respect to a stationary frame of reference S. Both the frames of reference have common origin. The time derivative of vector \(\bar{A}\) in S′ is :

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 \(-\bar{\omega} \times \bar{A}\).

Q: The plane of oscillation of a Foucault pendulum rotates :

The plane rotates 15° per hour at the pole.The precession rate of a Foucault pendulum is \(\Omega\sin\phi\), where \(\Omega = 360^{\circ}\) per sidereal day, about 15 degrees per hour, and \(\phi\) is the latitude. At a pole, \(\sin 90^{\circ} = 1\), so the plane turns a full circle in a day, 15 degrees per hour.At the equator \(\sin 0 = 0\), so there is no rotation. At latitude 60 degrees the rate is \(15 \times \sin 60^{\circ} \approx 13\) degrees per hour, not 7.5 (which would be \(15\sin 30^{\circ}\)) or 30.

Q: A particle is projected at an angle α with respect to the horizontal axis. Its path, observed in a frame of reference moving with a velocity equal to the horizontal component of its velocity, will be :

The path is a vertical straight line.The projectile's horizontal velocity \(u\cos\alpha\) is constant, since gravity acts only vertically. In a frame moving horizontally with exactly that velocity, the horizontal velocity becomes zero, and only the vertical motion remains: up with \(u\sin\alpha\), decelerating, then down. So the particle moves up and down along a vertical line.A line at angle \(\alpha\) to either axis would need a constant horizontal velocity relative to the frame, and a horizontal line would need no vertical motion.

Q: Consider S' a rotating frame rotating with constant angular velocity \(\omega\) with respect to stationary reference frame S (S and S' are having a common origin). If \(\bar{A}\) is any vector represented in S' frame, then \(\frac{d^{2}\bar{A}}{dt^{2}}\) in S frame is given by :

The correct expression is the one with \(+2\bar{\omega}\times\frac{d\bar{A}}{dt}\big|_{S'}\) and \(+\bar{\omega}\times\bar{\omega}\times\bar{A}\).For any vector, the rates of change in the two frames are related by \(\frac{d}{dt}\big|_{S} = \frac{d}{dt}\big|_{S'} + \bar{\omega}\times\). Applying this operator twice:\[ \begin{aligned} \ddot{\bar{A}}_{S} &= \Big(\tfrac{d}{dt}\Big|_{S'} + \bar{\omega}\times\Big)(\dot{\bar{A}}' + \bar{\omega}\times\bar{A}) \\ &= \ddot{\bar{A}}' + \dot{\bar{\omega}}\times\bar{A} + 2\bar{\omega}\times\dot{\bar{A}}' \\ &\quad + \bar{\omega}\times(\bar{\omega}\times\bar{A}) \end{aligned} \]where primes mean rates seen in \(S'\). The cross term appears twice, once from each application, which gives the factor 2 of the Coriolis term; the last term is the centripetal term. Both carry a plus sign here. The minus signs in the other options are what appear when the equation is solved for the acceleration in \(S'\), where the Coriolis and centrifugal terms move to the other side.

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?

Solid x and y axes of frame S with point P on the x axis, and dashed x-prime and y-prime axes of frame S-prime rotated anticlockwise about the common origin

A.\(F_{cen} = 0\) and \(F_{cor} = 0\)
B.\(F_{cen} \neq 0\) and \(F_{cor} \neq 0\) and \(F_{cen} = \frac{F_{cor}}{2}\)
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 \(F_{cen} = F_{cor}\)

Exam Year(s)

GATE Physics - 2026,

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.

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The period of oscillation of the plane of Foucault's pendulum is :

(\(\phi\) is the latitude)

A.Directly proportional to the latitude (\(\phi\))
B.Inversally proportional to sin \(\phi\)
C.Inversally proportional to cos \(\phi\)
D.Directly proportional to sin \(\phi\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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 \(\cos\phi\).

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A frame of reference S′ is rotating with constant angular velocity \(\omega\) with respect to a stationary frame of reference S. Both the frames of reference have common origin. The time derivative of vector \(\bar{A}\) in S′ is :

A.\(\frac{d'\bar{A}}{dt} = \bar{\omega} \times \bar{A}\)
B.\(\frac{d'\bar{A}}{dt} = \bar{\omega} \times \bar{A} + \frac{d\bar{A}}{dt}\)
C.\(\frac{d'\bar{A}}{dt} = -\bar{\omega} \times \bar{A} + \frac{d\bar{A}}{dt}\)
D.\(\frac{d'\bar{A}}{dt} = \frac{d\bar{A}}{dt}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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 \(-\bar{\omega} \times \bar{A}\).

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Consider two frames XYZ and xyz having common origin. The frame XYZ is fixed and the frame xyz is rotating with angular velocity (ω) with respect to the frame XYZ. Here \(\bar{\omega} = 2t\hat{x} - 5t^{2}\hat{z}\), where t represents time. For a given position vector (\(\bar{r}\)) defined as \(\bar{r} = 4t^{2}\hat{x} + 6t\hat{y} + t^{3}\hat{z}\) in the xyz coordinate frame, the velocities in frames XYZ and xyz for t = 1, will be respectively :

A.\(8\hat{x} - 6\hat{y} + 3\hat{z}\) and \(38\hat{x} + 16\hat{y} + 15\hat{z}\)
B.\(38\hat{x} - 16\hat{y} + 15\hat{z}\) and \(8\hat{x} + 6\hat{y} + 3\hat{z}\)
C.\(8\hat{x} + 6\hat{y} + 3\hat{z}\) and \(38\hat{x} - 16\hat{y} + 15\hat{z}\)
D.\(38\hat{x} + 16\hat{y} + 15\hat{z}\) and \(8\hat{x} - 6\hat{y} + 3\hat{z}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The velocities are \(38\hat{x} - 16\hat{y} + 15\hat{z}\) in XYZ and \(8\hat{x} + 6\hat{y} + 3\hat{z}\) in xyz.

In the rotating frame the velocity is the plain time derivative of \(\bar{r}\):

\[ \begin{aligned} \dot{\bar{r}} &= 8t\hat{x} + 6\hat{y} + 3t^{2}\hat{z} \\ &= 8\hat{x} + 6\hat{y} + 3\hat{z} \quad (t = 1) \end{aligned} \]

In the fixed frame add \(\bar{\omega} \times \bar{r}\), with \(\bar{\omega} = 2\hat{x} - 5\hat{z}\) and \(\bar{r} = 4\hat{x} + 6\hat{y} + \hat{z}\) at t = 1:

\[ \begin{aligned} \bar{\omega} \times \bar{r} &= (0 + 30)\hat{x} - (2 + 20)\hat{y} + (12 - 0)\hat{z} \\ &= 30\hat{x} - 22\hat{y} + 12\hat{z} \end{aligned} \]

Adding gives \(38\hat{x} - 16\hat{y} + 15\hat{z}\). The other options reverse the order of the two frames or carry the wrong sign on the \(\hat{y}\) components.

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The plane of oscillation of a Foucault pendulum rotates :

A.15° per hour at the equator
B.15° per hour at the pole
C.7.5° per hour at latitude 60°
D.30° per hour at latitude 60°

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The plane rotates 15° per hour at the pole.

The precession rate of a Foucault pendulum is \(\Omega\sin\phi\), where \(\Omega = 360^{\circ}\) per sidereal day, about 15 degrees per hour, and \(\phi\) is the latitude. At a pole, \(\sin 90^{\circ} = 1\), so the plane turns a full circle in a day, 15 degrees per hour.

At the equator \(\sin 0 = 0\), so there is no rotation. At latitude 60 degrees the rate is \(15 \times \sin 60^{\circ} \approx 13\) degrees per hour, not 7.5 (which would be \(15\sin 30^{\circ}\)) or 30.

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A particle is projected at an angle α with respect to the horizontal axis. Its path, observed in a frame of reference moving with a velocity equal to the horizontal component of its velocity, will be :

A.a straight line making an angle α with the horizontal
B.a horizontal straight line
C.a vertical straight line
D.a straight line making an angle α with the vertical

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The path is a vertical straight line.

The projectile's horizontal velocity \(u\cos\alpha\) is constant, since gravity acts only vertically. In a frame moving horizontally with exactly that velocity, the horizontal velocity becomes zero, and only the vertical motion remains: up with \(u\sin\alpha\), decelerating, then down. So the particle moves up and down along a vertical line.

A line at angle \(\alpha\) to either axis would need a constant horizontal velocity relative to the frame, and a horizontal line would need no vertical motion.

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Consider S' a rotating frame rotating with constant angular velocity \(\omega\) with respect to stationary reference frame S (S and S' are having a common origin). If \(\bar{A}\) is any vector represented in S' frame, then \(\frac{d^{2}\bar{A}}{dt^{2}}\) in S frame is given by :

A.\(\left.\frac{d^{2}\bar{A}}{dt^{2}}\right|_{S'}\) \(+ \left.\frac{d\bar{\omega}}{dt}\right|_{S'} \times \bar{A}\) \(+ 2\bar{\omega} \times \left.\frac{d\bar{A}}{dt}\right|_{S'}\) \(+ \bar{\omega} \times \bar{\omega} \times \bar{A}\)
B.\(\left.\frac{d^{2}\bar{A}}{dt^{2}}\right|_{S'}\) \(+ \left.\frac{d\bar{\omega}}{dt}\right|_{S'} \times \bar{A}\) \(+ 2\bar{\omega} \times \left.\frac{d\bar{A}}{dt}\right|_{S'}\) \(- \bar{\omega} \times \bar{\omega} \times \bar{A}\)
C.\(\left.\frac{d^{2}\bar{A}}{dt^{2}}\right|_{S'}\) \(+ \left.\frac{d\bar{\omega}}{dt}\right|_{S'} \times \bar{A}\) \(- 2\bar{\omega} \times \left.\frac{d\bar{A}}{dt}\right|_{S'}\) \(- \bar{\omega} \times \bar{\omega} \times \bar{A}\)
D.\(\left.\frac{d^{2}\bar{A}}{dt^{2}}\right|_{S'}\) \(+ \left.\frac{d\bar{\omega}}{dt}\right|_{S'} \times \bar{A}\) \(- 2\bar{\omega} \times \left.\frac{d\bar{A}}{dt}\right|_{S'}\) \(+ \bar{\omega} \times \bar{\omega} \times \bar{A}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The correct expression is the one with \(+2\bar{\omega}\times\frac{d\bar{A}}{dt}\big|_{S'}\) and \(+\bar{\omega}\times\bar{\omega}\times\bar{A}\).

For any vector, the rates of change in the two frames are related by \(\frac{d}{dt}\big|_{S} = \frac{d}{dt}\big|_{S'} + \bar{\omega}\times\). Applying this operator twice:

\[ \begin{aligned} \ddot{\bar{A}}_{S} &= \Big(\tfrac{d}{dt}\Big|_{S'} + \bar{\omega}\times\Big)(\dot{\bar{A}}' + \bar{\omega}\times\bar{A}) \\ &= \ddot{\bar{A}}' + \dot{\bar{\omega}}\times\bar{A} + 2\bar{\omega}\times\dot{\bar{A}}' \\ &\quad + \bar{\omega}\times(\bar{\omega}\times\bar{A}) \end{aligned} \]

where primes mean rates seen in \(S'\). The cross term appears twice, once from each application, which gives the factor 2 of the Coriolis term; the last term is the centripetal term. Both carry a plus sign here. The minus signs in the other options are what appear when the equation is solved for the acceleration in \(S'\), where the Coriolis and centrifugal terms move to the other side.

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