IIT JEE past question Mechanics 3

(a) Let the plank move by distance x, then the center of mass of cylinder moves by x/2.
Hence, by work energy principle
Fx=\dfrac{1}{2}m_2v_2^2+\dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}\cdot \dfrac{1}{2}m_1R^2\omega^2
Here, v₁ and v₂ are, respectively, the speeds of c.m of cylinder and the plank, and ω is the angular speed of the cylinder. Since
v_2=2v_1, \qquad \textrm{and}\quad \omega = \dfrac{v_1}{R}=\dfrac{v_2}{2R}
we get
Fx=\dfrac{1}{2}m_2v_2^2 + \dfrac{1}{2}m_1\dfrac{v_2^2}{4}+ \dfrac{1}{2}m_1\dfrac{v_2^2}{8}
i.e.
Fx=\dfrac{1}{2}v_2^2\left(m_2+\dfrac{3}{8}m_1\right)
Hence,
v_2=\sqrt{\dfrac{2Fx}{m_2+\frac{3}{8}m_1}}
Comparison with v=√2as gives the acceleration a₂ of the plank as
a_2=\dfrac{F}{m_2+\frac{3}{8}m_1}=\dfrac{8F}{8m_2+3m_1}
Hence, the acceleration of the center of mass of the cylinder
a_1=\dfrac{a_2}{2}=\dfrac{4F}{8m_2+3m_1}

(b) Let f_2 be the force of friction between the plank and the cylinder, then
F-f_2=m_2a_2, which give
f_2=F-m_2a_2=F-\dfrac{8m_2F}{8m_2+3m_1}=\dfrac{3m_1F}{8m_2+3m_1}
Let f₁ be the friction force between the cylinder and the ground, then f₁ must act forward. Hence, we get
f₁+f₂ = m₁a₁
from where we get
f_1=m_1a_1-f_2=\dfrac{4m_1F}{8m_2+3m_1}-\dfrac{3m_1F}{8m_2+3m_1}=\dfrac{m_1F}{8m_2+3m_1}

Magnitude and direction of friction

Friction at the top of the ball
The direction is opposite to the force applied.
Force of friction = Ï
₁m2g providing anti-clockwise motion

Friction at the bottom
The direction is to provide clockwise rotation.
Force of friction =Ï
₂(m1+m2)g