Question

A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets at the bottom of the plane first?
A. Disk
B. Sphere
C. Both reach at the same time
D. Depends upon their masses

Answer 1

Hint:since the inclined planes have same altitude and length it means total energy (rotational kinetic energy plus translational kinetic energy) will be same so the object which has more linear velocity at the bottom will reach first or the object which has more translational kinetic energy at the bottom of the inclined plane(assuming both objects disk and sphere performs pure rolling)

Complete step by step answer:
For an object rolling in an inclined plane,torque acting on it will be only due to friction.
Therefore,$\tau =fR$ where $f$ is friction and R is radius of the object.
Let the moment of inertia of the object be $I$. Then angular acceleration will be,
$$\begin{gathered} \alpha ={\displaystyle \frac{\tau }{I}} \\ \Rightarrow \alpha ={\displaystyle \frac{fR}{I}} \end{gathered}$$
And forces on the object will be gravitational force and friction
Acceleration along the incline will be $a={\displaystyle \frac{mg\sin \theta -f}{m}}$
For pure rolling,
$a=\alpha R$
$\Rightarrow {\displaystyle \frac{mg\sin \theta -f}{m}}={\displaystyle \frac{fR\times R}{I}}$
$\Rightarrow g\sin \theta ={\displaystyle \frac{f{R}^2}{I}}+{\displaystyle \frac{f}{m}}$
$$\begin{gathered} \Rightarrow f={\displaystyle \frac{mgI\sin \theta }{I+m{R}^2}} \\ \Rightarrow f={\displaystyle \frac{mg\sin \theta }{1+{\displaystyle \frac{m{R}^2}{I}}}} \end{gathered}$$
Once we know the value of friction we can find the value of acceleration.
Therefore,
$$\begin{gathered} a={\displaystyle \frac{mg\sin \theta -f}{m}} \\ \Rightarrow a={\displaystyle \frac{mg\sin \theta -{\displaystyle \frac{mgI\sin \theta }{I+m{R}^2}}}{m}} \\ \Rightarrow a=g\sin \theta (1-{\displaystyle \frac{I}{I+m{R}^2}}) \\ \Rightarrow a=g\sin \theta ({\displaystyle \frac{m{R}^2}{I+m{R}^2}}) \end{gathered}$$
If we write moment of inertia in terms of radius of gyration
i.e; $I=m{K}^2$
then value of acceleration $a={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{K^2}{R^2}}}}$
As we can see from the above equation,larger the value of K smaller the acceleration and vice versa.
For disc K is R and for sphere K is $\sqrt{{\displaystyle \frac{2}{5}}}$R. Since the value of K is smaller in sphere it means acceleration of sphere will be more therefore the sphere will reach first.

Hence Option B is correct.

Note:as here you can see acceleration does not depend upon the mass of the object it only depends upon value of K and value of K depends upon the geometry of the object. What it tells us is that no matter what the mass and size of an object is, as long as the geometry of the objects is the same it will take the same time to reach the ground in an inclined plane.