Question

Acceleration ($a$) of a body of radius R rolling down an inclined plane of inclination is given by?

Answer 1

Step 1: Given data

$K$ = Radius of gyration

$R$ = Radius of the object

$I$ = Inertia of the object

$N$ = Normal force acting on the body

$f$ = Frictional force

Step 2: Diagram

From the figure, $mg\sin \theta \&mg\cos \theta$=components of force due to body's weight ($mg$)

Where, $m$=mass of the object, $g$= acceleration due to gravity, and $\theta$ = angle of the inclined plane

Step 3: Calculating acceleration

From the figure, we can see that the sine component of the weight of the object, $mg\sin \theta$ acts towards left and the frictional force $f$ acts opposite to the direction of motion. Since the body is rolling down, the resultant force,

$ma=mg\sin \theta -f------\left(1\right)$

Here, the body is rolling due to friction, therefore net torque is given by,

Torque = force $\times$perpendicular distance,

$\tau =fR$

Also,

Torque = moment of inertia $\times$angular acceleration

$\tau =I\alpha$

$\therefore fR=mK²\times \alpha \left(\because I=mK²\right)-------\left(2\right)$

$K$ = radius of gyration

$R$ = radius of the object

$I$ = Inertia of the object

$N$ = Normal force acting on the body

$f$ = frictional force

In relation to linear and angular quantities, since there is no slipping

$v=R\omega$

$a=R\alpha$----(3)

$\alpha$=angular acceleration of the object

$v$=linear velocity of the object

$\omega$=angular velocity of the object

Solving 1,2 &3 simultaneously,

From 2 Value of $f=\frac{m{K}^2\alpha }{R}$substitute this value in 1

$$\begin{gathered} mg\sin \left(\theta \right)-\frac{m{k}^2\alpha }{R}=ma \\ g\sin \left(\theta \right)-\frac{K^2\alpha }{R}=a \end{gathered}$$-----(4)

From 3 $\alpha =\frac{a}{R}$substitute in 4

We get, $a=\frac{g\sin \theta }{1+{\displaystyle \frac{K^2}{R^2}}}$

Hence, the acceleration of a body rolling down an inclined plane is :$\mathit{a}\mathbf{=}\frac{\mathbf{g}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{[}\mathbf{1}\mathbf{+}{\displaystyle \frac{K^2}{R^2}}\mathbf{]}}$