Question

A frame of 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 ω is an example of a non-inertial frame of reference.

The relationship between the force ${\vec{F}}_{rot}$ experienced by a particle of mass m moving on the rotating disc and the force ${\vec{F}}_{in}$ experienced by the particle in an inertial frame of reference is

${\vec{F}}_{rot} = {\vec{F}}_{in} + 2 m ({\vec{v}}_{rot} \times \vec{ω}) + m (\vec{ω} \times \vec{r}) \times \vec{ω}$

where ${\vec{F}}_{rot}$ is the velocity of the particle in the rotating frame of reference and $\vec{r}$ 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 ω about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis $(\vec{ω} = ω \hat{k})$ . A small block of mass m is gently placed in the slot at $\vec{r} = (R / 2) \hat{i}$ at t = 0 and is constrained to move only along the Slot.

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Linked Question 1

The distance r of the block at time t is

$\frac{R}{4} (e^{ωt} + e^{- ωt})$

$\frac{R}{2} \cos 2 ωt$

$\frac{R}{4} (e^{2 ωt} + e^{- 2 ωt})$

$\frac{R}{2} cosωt$

$\frac{mdv}{dt} = {mω}^{2} x$

$v \frac{dv}{dx} = ω^{2} x$

$\int_{0}^{v} v dx = \int_{R / 2}^{x} ω^{2} x dx$

$\frac{v^{2}}{2} = {[\frac{ω^{2} x^{2}}{2}]}_{R / 2}^{x}$

$v^{2} = \frac{2 ω^{2}}{2} (x^{2} - \frac{R^{2}}{v})$

$v=ω \sqrt{x^{2} - {(\frac{R}{2})}^{2}}$

$\frac{dx}{dt} = ω \sqrt{x^{2} - {(\frac{R}{2})}^{2}}$

$\int_{R / 2}^{x} \frac{dx}{\sqrt{x^{2} - {(\frac{R}{2})}^{2}}} = ω \int_{0}^{t} dt$

$t = \frac{1}{ω} {[\cos h^{- 1} (\frac{x}{R / 2})]}_{R / 2}^{x}$

$= \frac{1}{ω} [\cos h^{- 1} (\frac{x}{R / 2}) - \cosh^{- 1}]$

$= \frac{1}{ω} \cos h^{- 1} (\frac{x}{R / 2})$

$\frac{R}{2} \cos h ωt = x$

$x = \frac{R}{2} \cos h ωt = \frac{R}{2} [\frac{e^{ωt} - e^{ωt}}{2}]$

Linked Question 2

The net reaction of the disc on the block is:

${mω}^{2} Rsinωt \hat{j} - mg \hat{k}$

$\frac{1}{2} {mω}^{2} R (e^{ωt} - e^{- ωt}) \hat{j} + mg \hat{k}$

$\frac{1}{2} {mω}^{2} R (e^{2 ωt} - e^{- 2 ωt}) \hat{j} + mg \hat{k}$

$- {mω}^{2} Rcosωt \hat{j} - mg \hat{k}$

Coriolis force $= 2 m \frac{dx}{dt} \times ω$

Net reaction $= \frac{2 mR}{4} ({ωe}^{ωt} - {ωe}^{- ωt}) ω \hat{j} + mg \hat{k}$

$= \frac{2 {mω}^{2} R}{4} (e^{ωt} - e^{- ωt}) \hat{j} + mg \hat{k}$

$= \frac{1}{2} {mω}^{2} R (e^{ωt} - e^{- ωt}) \hat{j} + mg \hat{k}$