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}

F - f = m2a2 ...(1)

(f-f2)R = Iα .........(2)

f + f2 = m1a1 ........(3)

a1 = Rα .........................(4)

a1 + a2 + Rα= 0 ...................(5)

PLS CHECK IF EQXNS ARE RITE SIR

ive taken alpha in anti clock dir and a1 and a2 rightward............

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