The following bodies, (1) a ring, (2) a disc, (3) a solid cylinder, (4) a solid sphere, of same mass $$m$$ and radius $$R$$ are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ________. [Mark the body as per their respective numbering given in the question]

We need to determine which of the four given bodies will reach the bottom of the inclined plane first when allowed to roll down simultaneously without slipping.

1. Relate Rolling Acceleration to the Moment of Inertia

When a body rolls down an inclined plane of angle $$\theta$$ without slipping, its linear acceleration ($$a$$) is given by the formula:

$$a = \frac{g \sin\theta}{1 + \frac{I}{mR^2}}$$

Where $$I$$ is the moment of inertia, $$m$$ is the mass, and $$R$$ is the radius of the body.

The body with the highest linear acceleration ($$a$$) will take the least amount of time to travel down the incline, meaning it will reach the bottom first. Looking at the formula, acceleration is inversely proportional to the factor $$\frac{I}{mR^2}$$. Therefore, the body with the smallest value of $$\frac{I}{mR^2}$$ will have the highest acceleration.

2. Analyze the Given Bodies

Let's calculate the value of $$\frac{I}{mR^2}$$ for each body mentioned on the  page:

3. Compare the Values

• The solid sphere has the lowest factor of $$\frac{I}{mR^2} = 0.40$$.
• Substituting this value into the acceleration formula gives it the maximum acceleration:

  $$a_{\text{sphere}} = \frac{g \sin\theta}{1 + 0.4} = \frac{5}{7}g \sin\theta \approx 0.714\, g \sin\theta$$

Conclusion

Since the solid sphere has the maximum linear acceleration, it will reach the bottom of the inclined plane first. This matches numbering 4.