A sphere rolls down on an inclined plane of inclination θ. What is the acceleration as the sphere reaches bottom

The correct option is 5g sinθ/7.

To find the acceleration of a solid sphere rolling down an inclined plane of inclination $\theta$ without slipping, we analyze both its translational and rotational motion.

Step 1: Equation of Translational Motion
Let $M$ be the mass of the sphere, $R$ be its radius, and $a$ be its linear acceleration down the incline.
The forces acting along the inclined plane are:
1. The component of gravitational force acting down the incline: $Mgsin\theta$
2. The static frictional force acting upwards along the incline (opposing the motion): $f$
Applying Newton's second law for translational motion:
$Mgsin\theta -f=Ma$ --- (Equation 1)

Step 2: Equation of Rotational Motion
The torque ($\tau$) about the center of mass is provided solely by the frictional force $f$:
$\tau =fR$
We also know that torque relates to angular acceleration ($\alpha$) by:
$\tau =I\alpha$
where $I$ is the moment of inertia of the sphere. Equating the two torque expressions gives:
$fR=I\alpha$ --- (Equation 2)

Step 3: Rolling Without Slipping Condition
For pure rolling (rolling without slipping), the relation between linear acceleration ($a$) and angular acceleration ($\alpha$) is:
$a=\alpha R\Rightarrow \alpha =\frac{a}{R}$
Substitute this value of $\alpha$ into Equation 2:
$fR=I\left(\frac{a}{R}\right)\Rightarrow f=\frac{Ia}{R^2}$ --- (Equation 3)

Step 4: Moment of Inertia of a Solid Sphere
The moment of inertia of a solid sphere about its central axis is:
$I=\frac{2}{5}M{R}^2$
Substitute this expression for $I$ into Equation 3:
$f=\frac{\left(\frac{2}{5}M{R}^2\right)a}{R^2}=\frac{2}{5}Ma$

Step 5: Calculate the Linear Acceleration
Substitute the expression for frictional force $f$ back into Equation 1:
$Mgsin\theta -\frac{2}{5}Ma=Ma$
Divide the entire equation by the mass $M$:
$gsin\theta -\frac{2}{5}a=a$
Rearrange to group the acceleration terms together:
$gsin\theta =a+\frac{2}{5}a$
$gsin\theta =\frac{7}{5}a$
Solving for $a$:
$a=\frac{5gsin\theta }{7}$

Thus, the acceleration of the sphere as it rolls down the inclined plane is $\frac{5}{7}gsin\theta$.