© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Forces Acting on the Body
Consider a round uniform body of mass $M$ and radius $R$ rolling down an inclined plane of angle $\theta$. The forces acting on the body are:
The component of gravitational force along the incline, $Mg \sin \theta$.
The normal force from the plane, perpendicular to the surface (which does not contribute to motion along the incline).
The frictional force $f$, acting up or down the plane, depending on the sense of rolling (it provides the torque needed for rotation, but does not dissipate energy if rolling is without slipping).
Step 2: Write the Equations for Translational and Rotational Motion
(a) Translational Motion: Along the inclined plane, the net force is:
$$
Mg \sin \theta - f = M a,
$$
where $a$ is the linear acceleration of the center of mass down the incline.
(b) Rotational Motion: The torque about the center of mass due to the frictional force $f$ is:
$$
\tau = f \times R = I \alpha,
$$
where $I$ is the moment of inertia of the body about its center of mass, $R$ is the radius, and $\alpha$ is the angular acceleration.
Step 3: Use the Rolling Without Slipping Condition
For rolling without slipping, $\alpha = \frac{a}{R}$. Thus, from the rotational equation,
$$
f \, R = I \,\alpha = I \,\frac{a}{R} \quad \Longrightarrow \quad f = \frac{I \, a}{R^2}.
$$
Step 4: Substitute Friction into the Translational Equation
Substitute $f = \frac{I \, a}{R^2}$ into the translational equation:
$$
Mg \sin \theta - \frac{I \, a}{R^2} = M a.
$$
Rearrange to solve for $a$:
$$
Mg \sin \theta = M a + \frac{I \, a}{R^2} = a \left( M + \frac{I}{R^2} \right).
$$
Therefore,
$$
a = \frac{Mg \sin \theta}{M + \frac{I}{R^2}}.
$$
We can factor out $M$ in the denominator:
$$
a = \frac{g \sin \theta}{1 + \frac{I}{M R^2}}.
$$
Step 5: Final Expression for the Acceleration
The linear acceleration $a$ of the rolling body is:
$$
\boxed{a = \frac{g \sin \theta}{1 + \frac{I}{M R^2}}.}
$$
