Question

A cylinder of mass \[{{M}_{c}}\] and sphere of mass ${{M}_{s}}$ are placed at points A and B of two inclines, respectively (see figure). If they roll on the incline without slipping such that their accelerations are the same, then the ratio $\dfrac{\sin {{\theta }_{c}}}{\sin {{\theta }_{s}}}$ is:
A. $\sqrt{\dfrac{8}{7}}$
B. $\sqrt{\dfrac{15}{14}}$
C. $\dfrac{8}{7}$
D. $\dfrac{15}{14}$

Answer 1

Hint: We have a cylinder and a sphere of different masses kept at different inclines rolling down without slipping and with the same acceleration. When there is no slipping we can say that there is a frictional force. We can consider a spherical body rolling down on an incline and find the equation for acceleration. Thus by equating the acceleration in this case we will get the solution.
Formula used:
$F=ma$
$\tau =I\alpha $
$I=m{{k}^{2}}$

Complete step by step answer:
In the question we are given a cylinder and a sphere of mass ${{M}_{c}}$ and ${{M}_{s}}$ respectively. These two bodies are placed on two different inclines surfaces and they roll on the incline without slipping.
Let us consider a circular body of mass ‘m’ on an inclined surface. The free body diagram of that body can be drawn as shown below.

The forces acting on the body are the frictional force, weight of the body and normal force.
Since the weight of the body is acting vertically downwards we resolve it into two components $mg\sin \theta $ and $mg\cos \theta $.
The acceleration of the body is ‘a’ and ‘$\alpha $’ is the angular acceleration.
Since the normal force and the cosine component of the weight of the body are equal and opposite, we can say that the force on the body is,
$F=mg\sin \theta -f$, where ‘f’ is the frictional force.
We also know that force can be given as, $F=ma$.
Equating these two forces we will get,
$\Rightarrow mg\sin \theta -f=ma$
By solving this we will get the frictional force as,
$\Rightarrow f=mg\sin \theta -ma$
$\Rightarrow f=m\left( g\sin \theta -a \right)$
We know that the torque about the centre is, $I\alpha $ where ‘I’ is the moment of inertia and ‘$\alpha $’is the angular acceleration.
We know that the moment of inertia about the centre is $I=m{{k}^{2}}$, where ‘k’ is the radius of gyration.
When there is no slipping, we know that
$a=\alpha R$, were ‘R’ is the radius of the body and ‘a’ is the acceleration
$\Rightarrow \dfrac{a}{R}=\alpha $
By applying the equation for torque about the centre of the body, we will get
$fR=I\alpha $
By applying the values for frictional force, moment of inertia and angular acceleration, we will get
$\Rightarrow m\left( g\sin \theta -a \right)R=m{{k}^{2}}\times \dfrac{a}{R}$
By solving this we can find the acceleration as,
$\Rightarrow \left( g\sin \theta -a \right)R={{k}^{2}}\times \dfrac{a}{R}$
$\Rightarrow g\sin \theta -a={{k}^{2}}\times \dfrac{a}{{{R}^{2}}}$
$\Rightarrow g\sin \theta =\dfrac{a{{k}^{2}}}{{{R}^{2}}}+a$
$\Rightarrow g\sin \theta =a\left( \dfrac{{{k}^{2}}}{{{R}^{2}}}+1 \right)$
$\Rightarrow a=\dfrac{g\sin \theta }{\left( \dfrac{{{k}^{2}}}{{{R}^{2}}}+1 \right)}$
Now in this case we have a cylinder and a sphere rolling down different inclined planes.
We know that the radius of gyration of cylinder is given as,
$k=\dfrac{1}{\sqrt{2}}R$
Therefore we can write,
$\dfrac{{{k}^{2}}}{{{R}^{2}}}=\dfrac{1}{2}$
Thus the acceleration of the cylinder will be,
$\Rightarrow {{a}_{c}}=\dfrac{g\sin {{\theta }_{c}}}{\left( \dfrac{1}{2}+1 \right)}$
$\Rightarrow {{a}_{c}}=\dfrac{g\sin {{\theta }_{c}}}{\left( \dfrac{3}{2} \right)}$
We have the radius of gyration of the sphere as,
$\Rightarrow k=\sqrt{\dfrac{2}{5}}R$
Therefore we get,
$\Rightarrow \dfrac{{{k}^{2}}}{{{R}^{2}}}=\dfrac{2}{5}$
Thus we get the acceleration of the sphere as,
$\Rightarrow {{a}_{s}}=\dfrac{g\sin {{\theta }_{s}}}{\left( \dfrac{2}{5}+1 \right)}$
$\Rightarrow {{a}_{s}}=\dfrac{g\sin {{\theta }_{s}}}{\left( \dfrac{7}{5} \right)}$
In the question it is given that the acceleration of both the sphere and the cylinder are the same.
Therefore we can write that,
${{a}_{c}}={{a}_{s}}$
$\Rightarrow \dfrac{g\sin {{\theta }_{c}}}{\left( \dfrac{3}{2} \right)}=\dfrac{g\sin {{\theta }_{s}}}{\left( \dfrac{7}{5} \right)}$
$\Rightarrow \dfrac{g\sin {{\theta }_{c}}}{g\sin {{\theta }_{s}}}=\dfrac{\left( \dfrac{3}{2} \right)}{\left( \dfrac{7}{5} \right)}$
$\Rightarrow \dfrac{\sin {{\theta }_{c}}}{\sin {{\theta }_{s}}}=\dfrac{15}{14}$
Therefore the ratio of $\dfrac{\sin {{\theta }_{c}}}{\sin {{\theta }_{s}}}$ is $\dfrac{15}{14}$.

So, the correct answer is “Option D”.

Note:
In this case to find the equation for acceleration we consider a circular body on an inclined plane and draw its free body diagram. We consider a circular body because the rolling surface of the cylinder and the sphere is circular.
A free body diagram is a diagram that gives us both the magnitude and direction of all the forces acting on a body. So, for similar questions constructing a free body diagram will make the question easy to solve.