Animated Solution for Physics - Laws of Motion: Comprehension Passage

A Deep Dive into Rotating Discs

The beauty of physics lies in how it allows us to shift our perspective. When we analyze a problem from the ground (an inertial frame), the math can sometimes become a tangled mess of complex trajectories. But if we step onto a rotating disc and view the world from its non-inertial perspective, the motion often simplifies into a straight line. However, this simplicity comes at a cost: we must invite the "ghosts" of classical mechanics—pseudo forces—to balance our equations.

In this classic JEE Advanced problem, we are tasked with tracking a block constrained to a smooth slot on a spinning disc. Let's break down the journey of this block, both radially and transversely.

The Radial Journey

Imagine you are standing on the disc, looking down at the slot. The slot is perfectly smooth, meaning there is absolutely no friction to impede the block's motion along the $x$-axis. Yet, the block accelerates outwards. Why? Because in this rotating frame, it experiences a centrifugal force.

The centrifugal force is given by the expression $m(\vec{\omega }\times \vec{r})\times \vec{\omega }$. Since the position vector $\vec{r}$ is purely along the $x$-axis and the angular velocity $\vec{\omega }$ is along the $z$-axis, this cross-product simplifies beautifully to a force acting purely outwards along the slot: $F_{\text{cf}}=m{\omega }^{2}r$.

Applying Newton's Second Law in this frame, we get:

To find the position as a function of time, we need to integrate. A clever trick here is to express acceleration as $v\frac{dv}{dr}$. This allows us to relate velocity directly to position before introducing time:

Integrating this from the initial position $R/2$ (where velocity is zero) to an arbitrary position $r$ yields the velocity profile:

Now, we substitute $v=\frac{dr}{dt}$ to find the time dependence. This leads to a standard logarithmic integral:

Evaluating this integral and applying the limits gives us a logarithmic equation. To isolate $r$, we convert the logarithm to an exponential form. A neat algebraic trick—squaring both sides after isolating the square root—eliminates the radical and allows the $r^2$ terms to cancel out perfectly. We finally arrive at the elegant expression for the radial distance:

The Transverse Reactions

Now, let's look at the forces acting perpendicular to the slot. The block is constrained; it cannot fly off the disc vertically, nor can it phase through the walls of the slot horizontally. The disc must exert normal reaction forces to enforce these constraints.

Vertically (along the $z$-axis), the floor of the slot pushes up to perfectly balance the downward pull of gravity. Thus, the vertical normal reaction is simply:

Horizontally (along the $y$-axis), things get more interesting. As the block moves radially outwards with velocity ${\vec{v}}_{\text{rot}}$, it cuts across the rotating frame. This motion triggers the Coriolis force, a bizarre pseudo-force that acts perpendicular to both the velocity and the axis of rotation.

The Coriolis force is given by $2m({\vec{v}}_{\text{rot}}\times \vec{\omega })$. With velocity along $\hat{i}$ and rotation along $\hat{k}$, the cross product $\hat{i}\times \hat{k}$ yields $-\hat{j}$. The Coriolis force pushes the block hard against the trailing wall of the slot:

To prevent the block from breaking through the wall, the slot must push back with an equal and opposite normal reaction, ${\vec{N}}_2$. By differentiating our position function $r(t)$ to find the velocity $v(t)$, and substituting it into the Coriolis equation, we find the magnitude of this side reaction:

The total net reaction of the disc on the block is the vector sum of these two independent normal forces. Combining them, we get our final, comprehensive answer:

This problem is a masterclass in non-inertial dynamics, seamlessly blending vector calculus, integration techniques, and a deep physical intuition of pseudo forces.