A uniform solid cylinder of mass $' m'$ rests on two horizontal planks. A light thread is wound on the cylinder. The hanging end of the thread is pulled vertically down with a constant force $F,$ which still does not bring any sliding of the cylinder. The coefficient of friction between the cylinder and planks is equal to $μ .$ If the maximum acceleration $a_{\max} (i n m / s^{2})$ of the axis of cylinder rolling over the planks is $a_{0},$ then what is the value of $\frac{3}{4} a_{0} ?$ (Assume $μ = 1 / 3$ and take $g = 10 m / s^{2}$ )

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 5.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Here friction on the cylinder through both the planks is static friction
Question Image
$N_{1} + N_{2} = F + m g ..... (1)$
$f_{1} + f_{2} = m a ..... (2)$
$F R - (f_{1} + f_{2}) R = \frac{m R^{2}}{2} α ............ (3)$
$a = α R ............ (4)$
from (2), (3) & (4)
$\Rightarrow \frac{f_{1} + f_{2}}{m} = \frac{F - (f_{1} + f_{2})}{m / 2}$
$\Rightarrow \frac{f_{1} + f_{2}}{2} = F - (f_{1} + f_{2})$
$\Rightarrow f_{1} + f_{2} = \frac{2 F}{3} ...... (3)$
Using eqn. (1) & (2)
$\frac{2 F}{3} = m a$
Using eqn. (1) & (3)
$\frac{2 F}{3} = (F + m g) μ$
$F = m g$
$∴ f_{1} + f_{2} = \frac{2 F}{3} = m a \Rightarrow a = \frac{2 g}{3} = \frac{20}{3}$