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 distance r of the block at time t is

R \over 2(e^\omega t + e^ - \omega t)

R \over 2(e^2\omega t + e^ - 2\omega t)

Rotational Motion PYQ — JEE Advanced 2016 · Shift 2 · Q51 with Solution

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Correct answer: THE CORRECT EXPRESSION SHOULD BE $R(T)=\DFRAC{R}{4}\LEFT(E^{\OMEGA T}+E^{-\OMEGA T}\RIGHT)$., IF FORCED TO CHOOSE FROM THE LISTED OPTIONS, OPTION C HAS THE RIGHT FUNCTIONAL FORM BUT WRONG COEFFICIENT.

Set up the rotating-frame equation

We are given

\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 moves in a smooth slot along the rotating diameter, so in the rotating frame its motion is only along the $X$ -axis.

Let

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

Since the slot is smooth, there is no force along the slot except the fictitious forces. The slot provides whatever constraint force is needed in the $Y$ -direction, but along $X$ we must have

F_{\text{rot},x}=0.

Compute the Coriolis force term

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

Since

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

we get

\vec v_{\text{rot}}\times \vec \omega = -\omega \dot x\,\hat j.

So the Coriolis term is purely in the $Y$ -direction and has no $X$ -component.

Compute the centrifugal term

First,

\vec \omega\times \vec r = \omega\hat k \times x\hat i = \omega x (\hat k\times \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 this term lies along the slot.

Equation of motion along the slot

Along $X$ ,

F_{\text{rot},x}=F_{\text{in},x}+m\omega^2 x.

But in the rotating frame,

F_{\text{rot},x}=m\ddot x.

Also, there is no real force along the smooth slot, so

F_{\text{in},x}=0.

Hence,

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

Therefore,

\ddot x = \omega^2 x.

Solve the differential equation

The general solution of

\ddot x = \omega^2 x

x(t)=A e^{\omega t}+B e^{-\omega t}.

Initial conditions:

At $t=0$ , the block is placed at $x(0)=\frac R2.$
It is gently placed, so relative to the slot its initial velocity is zero: $\dot x(0)=0.$

Now,

x(0)=A+B=\frac R2,

and

\dot x(t)=\omega A e^{\omega t}-\omega B e^{-\omega t},

\dot x(0)=\omega(A-B)=0 \implies A=B.

Thus,

A=B=\frac R4.

Therefore,

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

Since the question denotes the distance along the slot by $r$ ,

r(t)=\frac R4\left(e^{\omega t}+e^{-\omega t}\right)=\frac R2\cosh(\omega t).

Compare with options

A: $\dfrac R2\cos 2\omega t$ ❌
B: $\dfrac R2\cos \omega t$ ❌
C: $\dfrac R2\left(e^{\omega t}+e^{-\omega t}\right)$ ❌
D: $\dfrac R2\left(e^{2\omega t}+e^{-2\omega t}\right)$ ❌

Our derived result is

\boxed{r(t)=\frac R4\left(e^{\omega t}+e^{-\omega t}\right)}.

This does not match any option exactly. Option C has the correct exponent but is larger by a factor of 2.

So the stored answer C appears inconsistent with the initial condition $r(0)=R/2$ , because option C gives

r(0)=\frac R2(1+1)=R,

not $R/2$ .

Hence the likely intended correct expression is

\boxed{\frac R4\left(e^{\omega t}+e^{-\omega t}\right)}.

More from Rotational Motion

View written solutionFree

Set up the rotating-frame equation

We are given

\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 moves in a smooth slot along the rotating diameter, so in the rotating frame its motion is only along the $X$ -axis.

Let

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

Since the slot is smooth, there is no force along the slot except the fictitious forces. The slot provides whatever constraint force is needed in the $Y$ -direction, but along $X$ we must have

F_{\text{rot},x}=0.

Compute the Coriolis force term

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

Since

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

we get

\vec v_{\text{rot}}\times \vec \omega = -\omega \dot x\,\hat j.

So the Coriolis term is purely in the $Y$ -direction and has no $X$ -component.

Compute the centrifugal term

First,

\vec \omega\times \vec r = \omega\hat k \times x\hat i = \omega x (\hat k\times \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 this term lies along the slot.

Equation of motion along the slot

Along $X$ ,

F_{\text{rot},x}=F_{\text{in},x}+m\omega^2 x.

But in the rotating frame,

F_{\text{rot},x}=m\ddot x.

Also, there is no real force along the smooth slot, so

F_{\text{in},x}=0.

Hence,

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

Therefore,

\ddot x = \omega^2 x.

Solve the differential equation

The general solution of

\ddot x = \omega^2 x

x(t)=A e^{\omega t}+B e^{-\omega t}.

Initial conditions:

At $t=0$ , the block is placed at $x(0)=\frac R2.$
It is gently placed, so relative to the slot its initial velocity is zero: $\dot x(0)=0.$

Now,

x(0)=A+B=\frac R2,

and

\dot x(t)=\omega A e^{\omega t}-\omega B e^{-\omega t},

\dot x(0)=\omega(A-B)=0 \implies A=B.

Thus,

A=B=\frac R4.

Therefore,

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

Since the question denotes the distance along the slot by $r$ ,

r(t)=\frac R4\left(e^{\omega t}+e^{-\omega t}\right)=\frac R2\cosh(\omega t).

Compare with options

A: $\dfrac R2\cos 2\omega t$ ❌
B: $\dfrac R2\cos \omega t$ ❌
C: $\dfrac R2\left(e^{\omega t}+e^{-\omega t}\right)$ ❌
D: $\dfrac R2\left(e^{2\omega t}+e^{-2\omega t}\right)$ ❌

Our derived result is

\boxed{r(t)=\frac R4\left(e^{\omega t}+e^{-\omega t}\right)}.

This does not match any option exactly. Option C has the correct exponent but is larger by a factor of 2.

So the stored answer C appears inconsistent with the initial condition $r(0)=R/2$ , because option C gives

r(0)=\frac R2(1+1)=R,

not $R/2$ .

Hence the likely intended correct expression is

\boxed{\frac R4\left(e^{\omega t}+e^{-\omega t}\right)}.

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