A uniform disc is acted by two equal forces of magnitude `F`. One of them, acts tangentially to the disc, while other one is acting at the central point of the disc. The friction between disc surface and ground surface in `nF`. If `r` be the radius of the disc, then the value of `n` would be (in `N`)

Options

A. 0
B. 1.2
C. `2.0`
D. 3.2

Answer

To solve the problem step by step, we will analyze the forces acting on the uniform disc and apply the principles of rotational dynamics and translational motion.

### Step 1: Identify the Forces Acting on the Disc
The disc is acted upon by two equal forces of magnitude `F`:
- One force `F` acts tangentially to the disc.
- The other force `F` acts at the center of the disc.

Additionally, there is a friction force `f` acting at the point of contact with the ground, which is given as `nF`.

### Step 2: Establish the Equations of Motion
Let:
- `A` be the linear acceleration of the center of mass of the disc.
- `α` be the angular acceleration of the disc.

Since the point of contact with the ground has zero acceleration (it is rolling without slipping), we can relate the linear acceleration and angular acceleration as follows:
\[ A = \alpha r \]
where `r` is the radius of the disc.

### Step 3: Apply Newton's Second Law for Linear Motion
The net force acting on the disc in the horizontal direction can be expressed as:
\[ 2F - f = mA \]
where `m` is the mass of the disc.

### Step 4: Apply the Torque Equation
The torque about the center of mass due to the forces can be expressed as:
\[ \tau = F \cdot r + f \cdot r = I \alpha \]
where `I` is the moment of inertia of the disc. The moment of inertia for a uniform disc is given by:
\[ I = \frac{1}{2} m r^2 \]

Substituting the values, we have:
\[ F \cdot r + f \cdot r = \frac{1}{2} m r^2 \cdot \alpha \]

### Step 5: Substitute Angular Acceleration
From Step 2, we know:
\[ \alpha = \frac{A}{r} \]
Substituting this into the torque equation gives:
\[ F \cdot r + f \cdot r = \frac{1}{2} m r^2 \cdot \frac{A}{r} \]
This simplifies to:
\[ F + f = \frac{1}{2} m A \]

### Step 6: Substitute the Friction Force
Since the friction force is given as `f = nF`, we substitute this into the equation:
\[ F + nF = \frac{1}{2} m A \]
This simplifies to:
\[ (1 + n)F = \frac{1}{2} m A \]

### Step 7: Equate Linear Motion and Torque Equations
From the linear motion equation, we have:
\[ 2F - nF = mA \]
This can be rearranged to:
\[ (2 - n)F = mA \]

### Step 8: Relate the Two Equations
Now we have two equations:
1. \((1 + n)F = \frac{1}{2} m A\)
2. \((2 - n)F = mA\)

We can express `A` from the second equation:
\[ A = \frac{(2 - n)F}{m} \]

Substituting this expression for `A` into the first equation gives:
\[ (1 + n)F = \frac{1}{2} m \left(\frac{(2 - n)F}{m}\right) \]
This simplifies to:
\[ (1 + n)F = \frac{1}{2} (2 - n)F \]

### Step 9: Solve for `n`
Dividing both sides by `F` (assuming `F ≠ 0`):
\[ 1 + n = \frac{1}{2} (2 - n) \]
Multiplying through by 2 to eliminate the fraction:
\[ 2 + 2n = 2 - n \]
Rearranging gives:
\[ 3n = 0 \]
Thus:
\[ n = 0 \]

### Final Answer
The value of `n` is **0**.