Question

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $60^\circ $ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $\dfrac{{\left( {2 - \sqrt 3 } \right)}}{{\sqrt {10} s}}$ ,
Then the height of the top of the inclined plane, in meters. Is ___________. Take $g = 10m{s^{ - 2}}$.

Answer 1

Hint: Here, for the ring the axis is through its centre and not through its diameter and for the disc also its axis is passing through its centre and not from the diameter. Here, apply the formula for rolling from an inclined plane.

Complete step by step Solution:

The formula for acceleration of a rolling body from an inclined plane:

${a_c} = \dfrac{{g\sin \theta }}{{1 + \dfrac{{{I_c}}}{{M{R^2}}}}}$ ;

The moment of inertia of the ring is:

${a_{ring}} = \dfrac{{g\sin \theta }}{{1 + \dfrac{{M{R^2}}}{{M{R^2}}}}}$;
$ \Rightarrow {a_{ring}} = \dfrac{{g\sin \theta }}{{1 + \dfrac{{M{R^2}}}{{M{R^2}}}}}$;

Now, do the same for the Disc:

${a_c} = \dfrac{{g\sin \theta }}{{1 + \dfrac{{M{R^2}}}{{2M{R^2}}}}}$;

Take the LCM in the denominator and cancel the common factor:

$ \Rightarrow {a_c} = \dfrac{{2g\sin \theta }}{{2 + 1}}$;
$ \Rightarrow {a_c} = \dfrac{{2g\sin \theta }}{3}$;

Now, use the formula for kinematics and solve for the height:

$S = ut + \dfrac{1}{2}at_1^2$ ;

Here, $S = \dfrac{h}{{\sin \theta }}$ ;

$\dfrac{h}{{\sin \theta }} = u{t_1} + \dfrac{1}{2}at_1^2$;

Put the value of acceleration of the ring:

$\dfrac{h}{{\sin \theta }} = u{t_1} + \dfrac{1}{2}\left( {\dfrac{{g\sin \theta }}{2}} \right)t_1^2$;
$ \Rightarrow {t_1} = \sqrt {\dfrac{{4h}}{{g{{\sin }^2}\theta }}} = \sqrt {\dfrac{{16h}}{{3g}}} $;

Similarly, do for the Disc:

$\dfrac{h}{{\sin \theta }} = u{t_2} + \dfrac{1}{2}at_2^2$;

Put the value of acceleration of the Disc:

$ \Rightarrow \dfrac{h}{{\sin \theta }} = u{t_2} + \dfrac{1}{2}\left( {\dfrac{{g\sin \theta }}{2}} \right)t_2^2$;

$ \Rightarrow {t_2} = \sqrt {\dfrac{{3h}}{{g{{\sin }^2}\theta }}} = \sqrt {\dfrac{{4h}}{g}} $;

Now, we have been given time difference as:

${t_1} - {t_2} = \dfrac{{\left( {2 - \sqrt 3 } \right)}}{{\sqrt {10} s}}$;

Put the value of \[{t_1}\] and ${t_2}$ in the above equation and solve:

 $\sqrt {\dfrac{{16h}}{{3g}}} - \sqrt {\dfrac{{4h}}{g}} = \dfrac{{\left( {2 - \sqrt 3 } \right)}}{{\sqrt {10} s}}$;
\[ \Rightarrow \sqrt h \left( {\dfrac{4}{{\sqrt 3 }} - 2} \right) = 2 - \sqrt 3 \] ;

Do, the needed calculation:

$\sqrt h = \left( {\dfrac{{\left( {2 - \sqrt 3 } \right)\sqrt 3 }}{{\left( {4 - 2\sqrt 3 } \right)}}} \right)$ ;
$ \Rightarrow \sqrt h = \dfrac{{\sqrt 3 }}{2}$;

Take square on both the sides and calculate:

$ \Rightarrow h = \dfrac{3}{4} = 0.75m$;

Final Answer: The height of the top of the inclined plane, in meters. Is 0.75m.

Note:Here, we cannot solve the question just by applying the equation of kinematics. Indeed, the equation of kinematics is needed but we need to first find out the acceleration, it is counter intuitive because generally we take acceleration as gravitational acceleration but here since ring and disc is given so, we need to think outside of the box and find out the formula for acceleration in which moment of inertia is involved. After finding the moment of inertia put the value of acceleration in the distance equation of kinematics for both the ring and the disc and then write both the equations in terms of time and then subtract the value of both the times and equate it to the value of time difference given.