A circular disc with a groove along its diameter is placed horizontally. A block of mass 1 kg is pla

Q: Wait, why do we need to use a pseudo force here?

Because the disc is accelerating, it acts as a non-inertial frame of reference. Newton's laws only hold true in inertial (non-accelerating) frames. To 'fix' this and analyze the block's motion relative to the disc, we must introduce a pseudo force equal to $ma$, acting in the opposite direction of the frame's acceleration.

Q: Why can't we just use one normal reaction like we usually do?

In most standard problems, a block only rests on one surface (like a flat table or an incline). But here, the block is inside a **groove**. It is touching the bottom floor (due to gravity) AND the side wall (due to the pseudo force pushing it sideways). Each point of contact generates its own normal reaction and its own friction!

Q: How did we figure out that $\sin\theta = \frac{3}{5}$?

We were given $\cos\theta = \frac{4}{5}$. Remember the classic right triangle with sides 3, 4, and 5! Mathematically, you can use the identity $\sin^2\theta + \cos^2\theta = 1$. Plugging in $\cos\theta$, we get $\sin^2\theta = 1 - \frac{16}{25} = \frac{9}{25}$, which means $\sin\theta = \frac{3}{5}$.

Q: Why doesn't the block just fly out of the groove?

The pseudo force has a component perpendicular to the groove ($ma \sin\theta$). This component is pushing the block *into* the solid side wall of the groove. The wall is rigid, so it pushes back with an equal and opposite normal force ($N_2$), keeping the block perfectly constrained within the track.

Q: Is it possible to solve this problem from the ground frame instead?

Yes, absolutely! However, it is mathematically brutal. In the ground frame, the block's total acceleration is a vector sum of the disc's acceleration and the block's relative acceleration along the groove. By shifting to the non-inertial frame, we effectively 'stop' the disc, turning a complex 2D kinematics problem into a straightforward 1D force balance.

Question

A circular disc with a groove along its diameter is placed horizontally. A block of mass $1	ext{ kg}$ is placed as shown. The coefficient of friction between the block and all surfaces of groove in contact is $\mu = \frac{2}{5}$. The disc has an acceleration of $25	ext{ m/s}^2$. Find the acceleration of the block with respect to disc. Given $\cos 	heta = \frac{4}{5}$.

Sigma Percentile · Accepted Answer

### The Magic of Non-Inertial Frames

Imagine standing on a giant disc that is suddenly yanked to the left. What happens to you? You feel an invisible hand shoving you to the right! This "invisible hand" is what physicists call a **pseudo force**.

In this problem, we have a block sitting inside a groove on an accelerating disc. Instead of watching this chaos from the ground, we are going to jump onto the disc. By entering this **non-inertial frame**, we can treat the disc as if it were perfectly still, provided we apply a pseudo force of $ma$ to the block in the opposite direction of the disc's acceleration.

### Resolving the Forces

The pseudo force $ma$ acts horizontally to the right. However, the block is constrained to move along the groove, which is tilted at an angle $	heta$.

To understand the motion, we must resolve this pseudo force into two perpendicular components. The component acting along the groove is the driving force:

$$F_{	ext{drive}} = ma \cos	heta$$

The other component acts perpendicular to the groove, pushing the block hard against the side wall:

$$F_{	ext{press}} = ma \sin	heta$$

### The Double Friction Trap

Here is where many students fall into a trap. They assume there is only one normal reaction. But look closely at the geometry! The block is in contact with **two distinct surfaces**.

First, gravity pulls the block down against the floor of the groove. This gives us our first normal reaction:

$$N_1 = mg$$

Second, the perpendicular component of the pseudo force smashes the block against the side wall of the groove. This creates a second, horizontal normal reaction:

$$N_2 = ma \sin	heta$$

Because friction exists wherever there is contact and a normal force, we have **two frictional forces** fighting the block's motion. The friction from the floor is $f_1 = \mu N_1$, and the friction from the wall is $f_2 = \mu N_2$.

### The Master Equation

Now we can write Newton's Second Law for the block along the groove. The driving force pulls it forward, while both frictional forces drag it backward. If $a_r$ is the relative acceleration of the block, we have:

$$ma_r = ma \cos	heta - f_1 - f_2$$

Substituting our expressions for the frictional forces, the equation expands to:

$$ma_r = ma \cos	heta - \mu mg - \mu ma \sin	heta$$

Notice the sheer elegance of physics here—the mass $m$ appears in every single term! We can cancel it out entirely, proving that the relative acceleration is independent of the block's mass:

$$a_r = a \cos	heta - \mu g - \mu a \sin	heta$$

### Crunching the Numbers

With our master equation ready, it is time to execute the final calculation. We are given $a = 25$, $\mu = \frac{2}{5}$, and $g = 10$. We also know that $\cos	heta = \frac{4}{5}$.

Using the fundamental trigonometric identity, we can easily find the sine of the angle:

$$\sin	heta = \sqrt{1 - \cos^2	heta} = \frac{3}{5}$$

Now, we carefully substitute all these values into our acceleration equation:

$$a_r = 25\left(\frac{4}{5}ight) - \left(\frac{2}{5}ight)(10) - \left(\frac{2}{5}ight)(25)\left(\frac{3}{5}ight)$$

Let's break down the arithmetic step by step. The driving term becomes $20$. The floor friction term becomes $4$. The wall friction term becomes $6$.

$$a_r = 20 - 4 - 6$$

Subtracting the frictional decelerations from the driving acceleration leaves us with our final answer:

$$a_r = 10 \mathrm{\,m/s}^2$$

The block accelerates along the groove at exactly $10 \mathrm{\,m/s}^2$ relative to the disc.

Visualized Solution

$The Non-Inertial Frame$

$Components of Pseudo Force$

$Normal Reactions N_{1} and N_{2}$

$Frictional Forces f_{1} and f_{2}$

$Equation of Motion$

$Substituting Values$

$Final Calculation$

The Sigma Insight: Pseudo Force

The Magic of Non-Inertial Frames

Resolving the Forces

The Double Friction Trap

The Master Equation

Crunching the Numbers

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Non-Inertial Frame

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Components of Pseudo Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Normal Reactions N1​ and N2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Frictional Forces f1​ and f2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

The Sigma Insight: Pseudo Force

The Magic of Non-Inertial Frames

Resolving the Forces

The Double Friction Trap

The Master Equation

Crunching the Numbers

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