A frame of the reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity \omega is an example of a non-inertial frame of reference. The relationship between the force \overrightarrow F rot experienced by a particle of mass m moving on the rotating disc and the force \overrightarrow F in experienced by the particle in an inertial frame of reference is, \overrightarrow F rot = \overrightarrow F in + 2m ( \overrightarrow v rot \times\overrightarrow \omega ) + m ( \overrightarrow \omega \times\overrightarrow r ) \times\overrightarrow \omega , where, vrot is the velocity of the particle in the rotating frame of reference and r is the position vector of the particle with respect to the centre of the disc. Now, consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed \omega about its vertical axis through its centre. We assign a coordinate system with the origin at the centre of the disc, the X-axis along the slot, the Y-axis perpendicular to the slot and the Z-axis along the rotation axis ( \omega = \omega\widehat k ). A small block of mass m is gently placed in the slot at r = (R/2) \widehat i at t = 0 and is constrained to move only along the slot. The net reaction of the disc on the block is

1 \over 2m\omega ^2R(e^\omega t - e^ - \omega t)\widehat j + mg\widehat k

m\omega ^2R\sin \omega t\widehat j - mg\widehat k

1 \over 2m\omega ^2R(e^2\omega t - e^ - 2\omega t)\widehat j + mg\widehat k

- m\omega ^2R\cos \omega r\widehat j - mg\widehat k

Rotational Motion PYQ — JEE Advanced 2016 · Shift 2 · Q52 with Solution

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Correct answer: B

Set up the equation of motion in the rotating frame

In the rotating frame attached to the disc,

\vec F_{\text{rot}}=\vec F_{\text{in}}+2m(\vec v_{\text{rot}}\times \vec \omega)+m(\vec \omega\times \vec r)\times \vec \omega.

The block is constrained to move only along the slot, i.e. along the rotating $X$ -axis. So let

\vec r=x(t)\,\hat i, \qquad \vec v_{\text{rot}}=\dot x\,\hat i, \qquad \vec \omega=\omega\hat k.

The real forces acting on the block are:

gravity: $-mg\hat k$
reaction from the slot/disc: say $\vec N$

Since the slot is smooth, there is no reaction along $\hat i$ . Thus along $\hat i$ , only the inertial-force terms determine the motion.

Compute the pseudo-force terms

(a) Coriolis term

2m(\vec v_{\text{rot}}\times \vec \omega)=2m(\dot x\hat i\times \omega\hat k) =2m\omega\dot x(\hat i\times \hat k).

Using $\hat i\times \hat k=-\hat j$ ,

2m(\vec v_{\text{rot}}\times \vec \omega)=-2m\omega\dot x\hat j.

(b) Centrifugal term

First,

\vec \omega\times \vec r=\omega\hat k\times x\hat i=\omega x\hat j.

Then,

(\vec \omega\times \vec r)\times \vec \omega=(\omega x\hat j)\times (\omega\hat k) =\omega^2 x(\hat j\times \hat k)=\omega^2 x\hat i.

So,

m(\vec \omega\times \vec r)\times \vec \omega=m\omega^2 x\hat i.

Equation along the slot ( $\hat i$ -direction)

Since the block can move freely along the slot, the reaction has no $\hat i$ -component. Also, in the rotating frame, the block’s acceleration is just $\ddot x\hat i$ along the slot.

Hence along $\hat i$ ,

m\ddot x=m\omega^2 x.

So,

\ddot x=\omega^2 x.

Initial conditions:

at $t=0$ , the block is placed at $x(0)=R/2$
gently placed $\Rightarrow \dot x(0)=0$

General solution:

x=Ae^{\omega t}+Be^{-\omega t}.

Then

\dot x=\omega Ae^{\omega t}-\omega Be^{-\omega t}.

Using $\dot x(0)=0$ gives

\omega(A-B)=0 \Rightarrow A=B.

Using $x(0)=R/2$ gives

A+B=R/2 \Rightarrow 2A=R/2 \Rightarrow A=B=R/4.

Thus

x(t)=\frac R4\left(e^{\omega t}+e^{-\omega t}\right),

\dot x(t)=\frac{\omega R}{4}\left(e^{\omega t}-e^{-\omega t}\right).

Find the reaction force

The block has no motion in the $\hat j$ and $\hat k$ directions in the rotating frame, so the net force in those directions must vanish.

The real reaction force must balance:

Coriolis term in $\hat j$
gravity in $\hat k$

From the force relation,

\vec F_{\text{rot}}=\vec F_{\text{in}}+2m(\vec v_{\text{rot}}\times\vec\omega)+m(\vec\omega\times\vec r)\times\vec\omega.

In component form for $\hat j,\hat k$ , since acceleration in rotating frame is zero there,

0=N_j-2m\omega\dot x,

N_j=2m\omega\dot x.

Substitute $\dot x$ :

N_j=2m\omega\cdot \frac{\omega R}{4}\left(e^{\omega t}-e^{-\omega t}\right) =\frac12 m\omega^2R\left(e^{\omega t}-e^{-\omega t}\right).

In the vertical direction,

N_k-mg=0 \Rightarrow N_k=mg.

Hence the net reaction of the disc on the block is

\boxed{\vec N=\frac12 m\omega^2R\left(e^{\omega t}-e^{-\omega t}\right)\hat j+mg\hat k.}

Match with the options

This is exactly Option B.

\boxed{\text{B}}

More from Rotational Motion

View written solutionFree

Correct answer: B

Set up the equation of motion in the rotating frame

In the rotating frame attached to the disc,

\vec F_{\text{rot}}=\vec F_{\text{in}}+2m(\vec v_{\text{rot}}\times \vec \omega)+m(\vec \omega\times \vec r)\times \vec \omega.

The block is constrained to move only along the slot, i.e. along the rotating $X$ -axis. So let

\vec r=x(t)\,\hat i, \qquad \vec v_{\text{rot}}=\dot x\,\hat i, \qquad \vec \omega=\omega\hat k.

The real forces acting on the block are:

gravity: $-mg\hat k$
reaction from the slot/disc: say $\vec N$

Since the slot is smooth, there is no reaction along $\hat i$ . Thus along $\hat i$ , only the inertial-force terms determine the motion.

Compute the pseudo-force terms

(a) Coriolis term

2m(\vec v_{\text{rot}}\times \vec \omega)=2m(\dot x\hat i\times \omega\hat k) =2m\omega\dot x(\hat i\times \hat k).

Using $\hat i\times \hat k=-\hat j$ ,

2m(\vec v_{\text{rot}}\times \vec \omega)=-2m\omega\dot x\hat j.

(b) Centrifugal term

First,

\vec \omega\times \vec r=\omega\hat k\times x\hat i=\omega x\hat j.

Then,

(\vec \omega\times \vec r)\times \vec \omega=(\omega x\hat j)\times (\omega\hat k) =\omega^2 x(\hat j\times \hat k)=\omega^2 x\hat i.

So,

m(\vec \omega\times \vec r)\times \vec \omega=m\omega^2 x\hat i.

Equation along the slot ( $\hat i$ -direction)

Since the block can move freely along the slot, the reaction has no $\hat i$ -component. Also, in the rotating frame, the block’s acceleration is just $\ddot x\hat i$ along the slot.

Hence along $\hat i$ ,

m\ddot x=m\omega^2 x.

So,

\ddot x=\omega^2 x.

Initial conditions:

at $t=0$ , the block is placed at $x(0)=R/2$
gently placed $\Rightarrow \dot x(0)=0$

General solution:

x=Ae^{\omega t}+Be^{-\omega t}.

Then

\dot x=\omega Ae^{\omega t}-\omega Be^{-\omega t}.

Using $\dot x(0)=0$ gives

\omega(A-B)=0 \Rightarrow A=B.

Using $x(0)=R/2$ gives

A+B=R/2 \Rightarrow 2A=R/2 \Rightarrow A=B=R/4.

Thus

x(t)=\frac R4\left(e^{\omega t}+e^{-\omega t}\right),

\dot x(t)=\frac{\omega R}{4}\left(e^{\omega t}-e^{-\omega t}\right).

Find the reaction force

The block has no motion in the $\hat j$ and $\hat k$ directions in the rotating frame, so the net force in those directions must vanish.

The real reaction force must balance:

Coriolis term in $\hat j$
gravity in $\hat k$

From the force relation,

\vec F_{\text{rot}}=\vec F_{\text{in}}+2m(\vec v_{\text{rot}}\times\vec\omega)+m(\vec\omega\times\vec r)\times\vec\omega.

In component form for $\hat j,\hat k$ , since acceleration in rotating frame is zero there,

0=N_j-2m\omega\dot x,

N_j=2m\omega\dot x.

Substitute $\dot x$ :

N_j=2m\omega\cdot \frac{\omega R}{4}\left(e^{\omega t}-e^{-\omega t}\right) =\frac12 m\omega^2R\left(e^{\omega t}-e^{-\omega t}\right).

In the vertical direction,

N_k-mg=0 \Rightarrow N_k=mg.

Hence the net reaction of the disc on the block is

\boxed{\vec N=\frac12 m\omega^2R\left(e^{\omega t}-e^{-\omega t}\right)\hat j+mg\hat k.}

Match with the options

This is exactly Option B.

\boxed{\text{B}}

Rotational Motion question

Rotational Motion question

(a) Coriolis term

(b) Centrifugal term

More from Rotational Motion

Rotational Motion question

Rotational Motion question

(a) Coriolis term

(b) Centrifugal term

More from Rotational Motion