Question

Calculate the time taken by a uniform solid sphere and a disc of same mass and the same diameter to roll down through the distance from rest on a smooth inclined plane.
A.$15:14$
B.$\sqrt {15} :\sqrt {14}$
C.${15^2}:{14^2}$
D.$\sqrt {14} :\sqrt {15}$

Answer 1

We can use the formula of acceleration of an object rolling down from the inclined plane and use this acceleration in the displacement formula of the equation of motion. It gives the time taken by an object to roll down from the inclined plane.

Complete step by step solution:
There is a derived formula for acceleration of an object rolling down the inclined plane,
$a = \dfrac{{g\sin \theta }}{{1 + \dfrac{{{k^2}}}{{{R^2}}}}}$
Here, $g$ is acceleration due to gravity
$k$ is radius of gyration
$R$ is radius of object
We know from the equation of motions, the displacement of moving object can be determined by using following formula,
$s = ut + \dfrac{1}{2}a{t^2}$
Sphere and disc starts rolling from rest so initial velocity would be zero.
${l} s = 0 \times t + \dfrac{1}{2}a{t^2}\\\ s = \dfrac{1}{2}a{t^2}$
By rearranging the above equation we can write the expression in terms of time,
$t = \sqrt {\dfrac{{2s}}{a}}$
We can now put these values in individual ratios of time taken by solid sphere and time taken by disc to get the value of the ratio.
$t = \sqrt {\dfrac{{2s{R^2}}}{{g\sin \theta }}\left( {1 + \dfrac{{{k^2}}}{{{R^2}}}} \right)}$
The ratio of $\dfrac{{{k^2}}}{{{R^2}}}$for solid sphere is $\dfrac{2}{5}$ and for the disc is $\dfrac{1}{2}$.
We can now put these values in individual ratios of time taken by solid sphere and time taken by disc to get the value of the ratio.

$$\dfrac{{{t_s}}}{{{t_d}}} = \dfrac{{\sqrt {\dfrac{{2s{R^2}}}{{g\sin \theta }}\left( {1 + \dfrac{2}{5}} \right)} }}{{\sqrt {\dfrac{{2s{R^2}}}{{g\sin \theta }}\left( {1 + \dfrac{1}{2}} \right)} }}\\\ = \sqrt {\dfrac{{{7 {\left/ {\vphantom {7 5}} \right. } 5}}}{{{3 {\left/ {\vphantom {3 2}} \right. } 2}}}} \\\ = \sqrt {\dfrac{{14}}{{15}}}$$

Therefore, the time taken by a solid sphere and disc of the same mass and same diameter to roll from an inclined plane is $\sqrt {\dfrac{{14}}{{15}}}$. So from the given options we have to choose the correct option as D.

Note: It is important to remember the values of the ratio of $\dfrac{{{k^2}}}{{{R^2}}}$ for solid sphere as well as the disc. These types of problems become very easy when the ratio is known. Similarly, memorize the ratio for all the important shapes.