PARAGRAPH 1 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 $\omega$ is an example of a non-inertial frame of reference. The relationship between the force $\vec{F}_{\text{rot}}$ experienced by a particle of mass $m$ moving on the rotating disc and the force $\vec{F}_{\text{in}}$ experienced by the particle in an inertial frame of reference is \[ \vec{F}_{\mathrm{rot}}=\vec{F}_{\mathrm{in}}+2 m\left(\vec{v}_{\mathrm{rot}} \times \vec{\omega}\right)+m(\vec{\omega} \times \vec{r}) \times \vec{\omega}, \] where $\vec{v}_{\text{rot}}$ 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 $\omega$ 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 $(\vec{\omega}=\omega \hat{k})$. 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. The distance $r$ of the block at time $t$ is

(A) $\frac{R}{4}\left(e^{\omega t}+e^{-\omega t}\right)$

(B) $\frac{R}{2} \cos \omega t$

(C) $\frac{R}{4}\left(e^{2 \omega t}+e^{-2 \omega t}\right)$

(D) $\frac{R}{2} \cos 2 \omega t$

Laws of Motion — JEE Advanced · 2016 · 22 May 2016 · Paper 2 Question with Solution

Passage

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

F_{rot}

experienced by a particle of mass

m

moving on the rotating disc and the force

F_{in}

experienced by the particle in an inertial frame of reference is

F_{rot} = F_{in} + 2 m (v_{rot} \times ω) + m (ω \times r) \times ω,

where

v_{rot}

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

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

(ω = ω \hat{k})

. A small block of mass

m

is gently placed in the slot at

r = (R /2) \hat{i}

t = 0

and is constrained to move only along the slot.

PARAGRAPH 1

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

F_{rot}

experienced by a particle of mass

m

moving on the rotating disc and the force

F_{in}

experienced by the particle in an inertial frame of reference is

F_{rot} = F_{in} + 2 m (v_{rot} \times ω) + m (ω \times r) \times ω,

where

v_{rot}

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

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

(ω = ω \hat{k})

. A small block of mass

m

is gently placed in the slot at

r = (R /2) \hat{i}

t = 0

and is constrained to move only along the slot.

The distance

r

of the block at time

t

(A) $\frac{R}{4} (e^{ω t} + e^{- ω t})$ ✓ Correct
(B) $\frac{R}{2} cos ω t$
(C) $\frac{R}{4} (e^{2 ω t} + e^{- 2 ω t})$
(D) $\frac{R}{2} cos 2 ω t$

Answer: (A)

Solution

In the rotating frame, the block experiences centrifugal force

m ω^{2} r

outward.

Equation of motion:

m \frac{d v}{d t} = m ω^{2} r

Using

v \frac{d v}{d r} = ω^{2} r

and integrating from initial position

R /2

with

v = 0

\int_{0}^{v} v d v = ω^{2} \int_{R /2}^{r} r d r

\frac{v ^{2}}{2} = ω^{2} (\frac{r ^{2}}{2} - \frac{R ^{2}}{8})

v = ω r^{2} - \frac{R ^{2}}{4}

Separating variables:

\frac{d r}{r ^{2} - R ^{2} /4} = ω d t

Integrating:

cosh^{- 1} (\frac{2 r}{R}) = ω t + C

With

r (0) = R /2

C = 0

Therefore:

r = \frac{R}{4} (e^{ω t} + e^{- ω t})

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Single correct option

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Laws of Motion

Solution

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