Courses
Courses for Kids
Free study material
Offline Centres
More

A disc has a radius $20cm$ and a mass half kg found to be rolling on an inclined plane. What will be the force of friction so that the disc will perform pure rolling?
seo images

$\begin{align}
  & A.\dfrac{5\sqrt{2}}{3}N \\
 & B.\dfrac{5}{3\sqrt{2}}N \\
 & C.\dfrac{5}{\sqrt{2}}N \\
 & D.\dfrac{5}{2\sqrt{3}}N \\
\end{align}$

seo-qna
Last updated date: 01st Mar 2024
Total views: 361.2k
Views today: 7.61k
IVSAT 2024
Answer
VerifiedVerified
361.2k+ views
Hint: From the diagram, derive the equation of the balancing of the force. The torque of the centre of mass can be found to be equivalent to the product of the moment of inertia of the centre of mass and the angular acceleration of the body. Rearrange the equation in terms of friction. Substitute the torque in this and substitute the values. This will help you in answering this question.

Complete step by step answer:
seo images

It has been mentioned that the radius of the disk has been given as,
$r=20cm$
Mass of the body has been mentioned as,
$m=\dfrac{1}{2}Kg$
According to the diagram mentioned, we can write that,
$mg\sin \theta -f=m{{a}_{cm}}$…….. (1)
Where $f$be the force of friction.
The torque of the centre of mass can be found to be equivalent to the product of the moment of inertia of the centre of mass and the angular acceleration of the body. That is we can write that,
${{\tau }_{cm}}={{I}_{cm}}\cdot \alpha $
And also we can write that,
\[{{f}_{R}}=\dfrac{M{{R}^{2}}}{2}\times \left( \dfrac{{{a}_{cm}}}{{{R}^{2}}} \right)\]
Simplifying this equation can be written as,
\[{{f}_{R}}=\dfrac{M}{2}\times {{a}_{cm}}\]
Rearranging this equation can be written as,
\[{{a}_{cm}}=\dfrac{2f}{M}\]
Substituting this in equation (1) can be shown as,
\[mg\sin \theta -f=M\left( \dfrac{2f}{M} \right)\]
From this the force of friction can be found to be as,
\[f=\dfrac{mg\sin \theta }{3}\]
The angle of inclination has been mentioned as,
\[\theta =45{}^\circ \]
Substituting all these values in it will give,
\[f=\dfrac{1\times 10\times 1}{2\times 3\times \sqrt{2}}=\dfrac{5}{3\sqrt{2}}N\]
Hence the force of friction has been found to be,
\[f=\dfrac{5}{3\sqrt{2}}N\]

So, the correct answer is “Option B”.

Note: Centre of mass is the imaginary point of a body or a particle which is considered to be the point where the whole mass of the body is found to be concentrated. All the properties of the body will be shown by this particle. Frictional force is the opposition provided due to the surface body interactions.
Recently Updated Pages
Trending doubts