An inclined plane makes an angle of `60^(@)` with horizontal. A disc rolling down this inclined plane without slipping has a linear acceleration equal to

Options

A. `(g)/(3)`
B. `(3)/(4)g`
C. `(g)/sqrt3`
D. `(g)/(2)`

Answer

To find the linear acceleration of a disc rolling down an inclined plane at an angle of \(60^\circ\) with the horizontal, we can follow these steps:

### Step 1: Identify the formula for linear acceleration of a rolling body
The linear acceleration \(a\) of a rolling body can be calculated using the formula:
\[ a = \frac{g \sin \theta}{1 + \frac{k^2}{r^2}} \]
where:
- \(g\) is the acceleration due to gravity,
- \(\theta\) is the angle of inclination,
- \(k\) is the radius of gyration,
- \(r\) is the radius of the disc.

### Step 2: Calculate the radius of gyration for a disc
The radius of gyration \(k\) is given by:
\[ k = \sqrt{\frac{I}{m}} \]
For a disc, the moment of inertia \(I\) is:
\[ I = \frac{1}{2} m r^2 \]
Substituting this into the formula for \(k\):
\[ k = \sqrt{\frac{\frac{1}{2} m r^2}{m}} = \sqrt{\frac{1}{2} r^2} = \frac{r}{\sqrt{2}} \]

### Step 3: Substitute \(k\) into the acceleration formula
Now, substituting \(k = \frac{r}{\sqrt{2}}\) into the acceleration formula:
\[ a = \frac{g \sin \theta}{1 + \frac{\left(\frac{r}{\sqrt{2}}\right)^2}{r^2}} = \frac{g \sin \theta}{1 + \frac{r^2/2}{r^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]

### Step 4: Calculate \(\sin 60^\circ\)
We know that:
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

### Step 5: Substitute \(\sin 60^\circ\) into the acceleration formula
Now substituting \(\sin 60^\circ\) into the equation for \(a\):
\[ a = \frac{2g \cdot \frac{\sqrt{3}}{2}}{3} = \frac{g \sqrt{3}}{3} \]

### Step 6: Final result
Thus, the linear acceleration of the disc rolling down the inclined plane is:
\[ a = \frac{g \sqrt{3}}{3} \]

### Step 7: Simplify the expression
This can also be expressed as:
\[ a = \frac{g}{\sqrt{3}} \]

### Conclusion
The final answer is:
\[ \text{The linear acceleration of the disc is } \frac{g}{\sqrt{3}}. \]

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