Question

two discs having masses in the ratio 1:3 and radii in the ratio 1:2 roll down without slipping one by one from the top of a inclined plane of heught 10m the ratio of their linrar velocities on reaching the ground is

Answer 1

When a body starts to roll along an inclined plane without slipping from a height h, it has only potential energy at the height.

• $\displaystyle\sf {E_1=mgh}$

When it reaches ground, it has translational as well as rotational kinetic energy only.

• $\displaystyle\sf{E_2=\dfrac {1}{2}\,mv^2\left (1+\dfrac {K^2}{R^2}\right)}$

By energy conservation,

$\displaystyle\sf{\longrightarrow E_1=E_2}$

$\displaystyle\sf{\longrightarrow mgh=\dfrac {1}{2}\,mv^2\left (1+\dfrac {K^2}{R^2}\right)}$

$\displaystyle\sf{\longrightarrow v=\sqrt{\dfrac{2gh}{\left (1+\dfrac {K^2}{R^2}\right)}}}$

This equation implies the velocity of the body at the bottom is independent of mass and individual radius, but depends only on height $\displaystyle\sf {h}$ and value of $\displaystyle\sf {\dfrac {K^2}{R^2}.}$

In the question two 'discs' are rolling so both have same $\displaystyle\sf {\dfrac {K^2}{R^2}}$ value (i.e., 1/2).

Both discs are releasing from 'same height' so $\displaystyle\sf {h}$ is same.

These two are enough to imply that both discs reach ground with same linear velocity.

Hence the ratio of the linear velocities is 1:1.