The net reaction of the disc on the block 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 net reaction of the disc on the block is [2016]

Smart Achievers · Accepted Answer

(2)

${\vec{F}}_{rot} = {\vec{F}}_{in} + 2 m (V_{rot} \hat{i}) \times ω \hat{k} + m (ω \hat{k} \times r \hat{i}) \times ω \hat{k}$

$∴$ $m r ω^{2} \hat{i} = {\vec{F}}_{in} + 2 m V_{rot} ω (- \hat{j}) + m ω^{2} r \hat{i}$

${\vec{F}}_{in} = m r V_{rot} ω \hat{j} . . . (i)$

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

$∴$ $\frac{d r}{d t} = V_{r} = \frac{R}{4} (ω e^{ω t} - ω e^{- ω t}) . . . (i i)$

$From (i) and (ii)$

${\vec{F}}_{in} = 2 m \frac{R ω}{4} (e^{ω t} - e^{- ω t}) ω \hat{j}$

$∴$ ${\vec{F}}_{in} = \frac{m R ω^{2}}{2} (e^{ω t} - e^{- ω t}) \hat{j}$

$∴$ ${\vec{F}}_{reaction} = \frac{m R ω^{2}}{2} (e^{ω t} - e^{- ω t}) \hat{j} + m g \hat{k}$

Solution

1 $\frac{1}{2} m ω^{2} R (e^{2 ω t} - e^{- 2 ω t}) \hat{j} + m g \hat{k}$

2 $\frac{1}{2} m ω^{2} R (e^{ω t} - e^{- ω t}) \hat{j} + m g \hat{k}$

3 $- m ω^{2} R \cos ω t \hat{j} - m g \hat{k}$

4 $m ω^{2} R \sin ω t \hat{j} - m g \hat{k}$

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Solution

1 12mω2R(e2ωt-e-2ωt)j^+mgk^

2 12mω2R(eωt-e-ωt)j^+mgk^

3 -mω2Rcosωt j^-mgk^

4 mω2Rsinωt j^-mgk^

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1 $\frac{1}{2} m ω^{2} R (e^{2 ω t} - e^{- 2 ω t}) \hat{j} + m g \hat{k}$

2 $\frac{1}{2} m ω^{2} R (e^{ω t} - e^{- ω t}) \hat{j} + m g \hat{k}$

3 $- m ω^{2} R \cos ω t \hat{j} - m g \hat{k}$

4 $m ω^{2} R \sin ω t \hat{j} - m g \hat{k}$