The distance r of the block at time t is [2016]

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 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 ${\vec{v}}_{rot}$ 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 $\vec{r} = (\frac{R}{2}) \hat{i}$ at $t = 0$ and is constrained to move only along the slot.

Q. The distance $r$ of the block at time $t$ is [2016]

Smart Achievers · Accepted Answer

(1)

Force on the block along slot ${\vec{F}}_{rot}$

$∴$ $\int_{0}^{v} V d v = \int_{r / 2}^{r} ω^{2} r d r \Rightarrow v = ω \sqrt{r^{2} - \frac{R^{2}}{4}} = \frac{d r}{d t}$

$ω$ $\int_{R / 4}^{r} \frac{d r}{\sqrt{r^{2} - \frac{R^{2}}{4}}} = \int_{0}^{t} ω d t$

$t = 0$

$or r^{2} - \frac{R^{2}}{4} = \frac{R^{2}}{4} e^{2 ω t} + r^{2} - 2 r \cdot \frac{R}{2} e^{ω t}$

$∴$ $r = \frac{R}{4} (e^{ω t} + e^{- ω t})$

Solution

1 $\frac{R}{4} (e^{ω t} + e^{- ω t})$

2 $\frac{R}{2} \cos ω t$

3 $\frac{R}{4} (e^{2 ω t} + e^{- 2 ω t})$

4 $\frac{R}{2} \cos 2 ω t$

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Solution

1 R4(eωt+e-ωt)

2 R2cosωt

3 R4(e2ωt+e-2ωt)

4 R2cos2ωt

Ans. (1) Force on the block along slot =mrω2=ma=mvdvdr ∴ ∫0vV dv=∫r/2rω2r dr⇒v=ωr2-R24=drdt ∴∫R/4rdrr2-R24=∫0tω dt On solving we get r+r2-R24=R2eωt or r2-R24=R24e2ωt+r2-2r·R2eωt ∴ r=R4(eωt+e-ωt)

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1 $\frac{R}{4} (e^{ω t} + e^{- ω t})$

2 $\frac{R}{2} \cos ω t$

3 $\frac{R}{4} (e^{2 ω t} + e^{- 2 ω t})$

4 $\frac{R}{2} \cos 2 ω t$