A frame of reference that is accelerated with respect to an inertial frame of reference is called a

Q: Why is the Coriolis force directed along the negative Y-axis?

The Coriolis force is given by $-2m(\vec{\omega} \times \mathbf{v}_{\text{rot}})$. Here, $\vec{\omega}$ is in the $+\hat{\mathbf{k}}$ direction and $\mathbf{v}_{\text{rot}}$ is in the $+\hat{\mathbf{i}}$ direction. Their cross product is $\hat{\mathbf{j}}$, so the force is in the $-\hat{\mathbf{j}}$ direction.

Q: Why doesn't the centrifugal force have a Y-component?

The centrifugal force is $m\omega^2 \mathbf{r}$. Since the block is constrained to move along the slot which lies on the X-axis, its position vector $\mathbf{r}$ only has an X-component. Thus, the centrifugal force is purely radial.

Q: How did we solve the integral $\int \frac{dr}{\sqrt{r^2 - a^2}}$?

This is a standard integral in calculus. The formula is $\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln|x + \sqrt{x^2 - a^2}| + C$. We applied this formula and evaluated it from the initial position $R/2$ to $r$.

Q: Why is the real force along the X-axis zero?

The problem states that the slot is 'smooth', which means there is no friction. The normal reactions from the slot walls are perpendicular to the slot (along Y and Z axes). Therefore, no real force acts along the X-axis.

Q: What is the physical meaning of the exponential terms in the answer?

The exponential terms indicate that the block's distance from the center grows exponentially over time. This makes sense because as the block moves further out, the centrifugal force increases, causing it to accelerate even faster.

Question

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 $\omega$ is an example of a non-inertial frame of reference. The relationship between the force $\mathbf{F}_{	ext{rot}}$ experienced by a particle of mass $m$ moving on the rotating disc and the force $\mathbf{F}_{	ext{in}}$ experienced by the particle in an inertial frame of reference is,

$\mathbf{F}_{	ext{rot}} = \mathbf{F}_{	ext{in}} + 2m(\mathbf{v}_{	ext{rot}} 	imes \vec{\omega}) + m(\vec{\omega} 	imes \mathbf{r}) 	imes \vec{\omega}$,

where, $\mathbf{v}_{	ext{rot}}$ is the velocity of the particle in the rotating frame of reference and $\mathbf{r}$ 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 ($\vec{\omega} = \omega \hat{\mathbf{k}}$). A small block of mass $m$ is gently placed in the slot at $\mathbf{r} = (R/2)\hat{\mathbf{i}}$ at $t = 0$ and is constrained to move only along the slot.

Question 1:

Sigma Percentile · Accepted Answer

### The Magic of Rotating Frames

Have you ever tried walking in a straight line on a spinning merry-go-round? It feels like an invisible hand is pushing you sideways. This invisible hand is what physicists call a "pseudo force." When we analyze motion from the perspective of an accelerating or rotating observer, Newton's laws of motion don't work in their standard form. To fix this, we introduce pseudo forces.

In this problem, we are dealing with a block sliding in a smooth slot on a rotating disc. The passage gives us the master equation for the force in the rotating frame:

$\mathbf{F}_{	ext{rot}} = \mathbf{F}_{	ext{in}} + 2m(\mathbf{v}_{	ext{rot}} 	imes \vec{\omega}) + m(\vec{\omega} 	imes \mathbf{r}) 	imes \vec{\omega}$

This equation might look intimidating, but it is simply Newton's second law adapted for a rotating world. The term $\mathbf{F}_{	ext{in}}$ is the real, physical force acting on the block. The other two terms are the pseudo forces: the Coriolis force and the centrifugal force.

### Unleashing the Pseudo Forces

Let's break down these pseudo forces. The block is constrained to move along the slot, which we've aligned with the X-axis. So, its position vector is $\mathbf{r} = r \hat{\mathbf{i}}$, and its velocity in the rotating frame is $\mathbf{v}_{	ext{rot}} = v_r \hat{\mathbf{i}}$. The disc rotates about the Z-axis, so $\vec{\omega} = \omega \hat{\mathbf{k}}$.

First, the centrifugal force. This is the force that pushes you outward on a merry-go-round. Mathematically, it is:

$m(\vec{\omega} 	imes \mathbf{r}) 	imes \vec{\omega} = m(\omega \hat{\mathbf{k}} 	imes r \hat{\mathbf{i}}) 	imes \omega \hat{\mathbf{k}} = m(\omega r \hat{\mathbf{j}}) 	imes \omega \hat{\mathbf{k}} = m\omega^2 r \hat{\mathbf{i}}$

Notice that this force points purely in the positive X-direction. It acts to push the block radially outward along the slot.

Next, the Coriolis force. This force acts on objects moving within a rotating frame.

$2m(\mathbf{v}_{	ext{rot}} 	imes \vec{\omega}) = 2m(v_r \hat{\mathbf{i}} 	imes \omega \hat{\mathbf{k}}) = -2m v_r \omega \hat{\mathbf{j}}$

This force points in the negative Y-direction. It tries to push the block sideways, against the wall of the slot.

### The Radial Sprint: Solving for Position

Now, let's look at the motion along the X-axis. The slot is perfectly smooth, meaning there is no friction. The only force acting along the X-axis is the centrifugal force. Using Newton's second law in the rotating frame:

$m a_x = m\omega^2 r$

We can rewrite the acceleration $a_x$ using the chain rule as $v_r \frac{dv_r}{dr}$:

$v_r \frac{dv_r}{dr} = \omega^2 r$

This is a separable differential equation. We can integrate both sides. The block starts at $r = R/2$ from rest ($v_r = 0$).

$\int_0^{v_r} v dv = \int_{R/2}^r \omega^2 r dr$

Evaluating the integrals, we get:

$\frac{v_r^2}{2} = \frac{\omega^2}{2} \left( r^2 - \frac{R^2}{4} ight)$

Taking the square root gives us the velocity as a function of position:

$v_r = \omega \sqrt{r^2 - \frac{R^2}{4}}$

To find the position as a function of time, we replace $v_r$ with $\frac{dr}{dt}$ and separate the variables again:

$\int_{R/2}^r \frac{dr}{\sqrt{r^2 - \frac{R^2}{4}}} = \int_0^t \omega dt$

This is a standard logarithmic integral. The solution is:

$\left[ \ln \left( r + \sqrt{r^2 - \frac{R^2}{4}} ight) ight]_{R/2}^r = \omega t$

Substituting the limits:

$\ln \left( \frac{r + \sqrt{r^2 - \frac{R^2}{4}}}{R/2} ight) = \omega t$

Taking the exponential of both sides:

$r + \sqrt{r^2 - \frac{R^2}{4}} = \frac{R}{2} e^{\omega t}$

To isolate $r$, we move it to the right side and square both sides:

$r^2 - \frac{R^2}{4} = \frac{R^2}{4} e^{2\omega t} - R r e^{\omega t} + r^2$

The $r^2$ terms cancel out beautifully. Solving for $r$, we find:

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

This elegant result tells us that the block's distance from the center grows exponentially over time. This is the answer to the first question!

### The Sideways Push: Finding the Reaction Force

Now, let's tackle the second question: finding the net reaction force from the disc on the block. The block is constrained to move only along the slot, which means it cannot accelerate in the Y or Z directions.

Along the Y-axis, the net force must be zero. The forces acting in the Y-direction are the real normal reaction from the side of the slot, $(\mathbf{F}_{	ext{in}})_y$, and the Coriolis force.

$(\mathbf{F}_{	ext{in}})_y - 2m v_r \omega = 0$

$(\mathbf{F}_{	ext{in}})_y = 2m v_r \omega$

To find this force, we need the velocity $v_r$. We can find it by differentiating our position equation with respect to time:

$v_r = \frac{dr}{dt} = \frac{R\omega}{4} (e^{\omega t} - e^{-\omega t})$

Substituting this into our force equation:

$(\mathbf{F}_{	ext{in}})_y = 2m \omega \left[ \frac{R\omega}{4} (e^{\omega t} - e^{-\omega t}) ight] = \frac{1}{2} m\omega^2 R (e^{\omega t} - e^{-\omega t})$

This is the sideways push exerted by the wall of the slot to keep the block moving in a straight line within the rotating frame.

Finally, we must consider the Z-axis. The block is resting on the disc, so gravity is pulling it down, and the bottom of the slot is pushing it up. Since there is no vertical acceleration:

$(\mathbf{F}_{	ext{in}})_z - mg = 0$

$(\mathbf{F}_{	ext{in}})_z = mg$

Combining the Y and Z components, we get the total reaction force exerted by the disc on the block:

$\mathbf{F}_{	ext{reaction}} = \frac{1}{2} m\omega^2 R (e^{\omega t} - e^{-\omega t}) \hat{\mathbf{j}} + mg \hat{\mathbf{k}}$

This perfectly matches our options. By carefully applying the principles of non-inertial frames and breaking the problem down into its directional components, we've unraveled the complex motion of the block.

Animated Solution for Physics - Laws of Motion: Comprehension Passage

Select Answer:

Visualized Solution

The Rotating Frame

Equation of Motion in Rotating Frame

Identifying Pseudo Forces

Motion Along the Slot (X-axis)

Integrating for Velocity

Setting up the Time Integral

Solving the Integral

Position as a Function of Time

Forces Along the Y-axis

Calculating the Y-Reaction

Forces Along the Z-axis

Net Reaction Force

The Sigma Insight: Pseudo Force

The Magic of Rotating Frames

Unleashing the Pseudo Forces

The Radial Sprint

The Sideways Push

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