# Friction between two disks

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Two discs A and B touch each other as in figure. A rope tightly wound on A is pulled down at 2 m/s$^2$ . Find the friction force between A and B if slipping is absent

My try

But the answer given in book is 2N

edited Dec 20, 2016
totally wrong !!!

The answer of $2m/s^2$ is an acceleration not a force, so it cannot be the friction force.

The rope attached to disk A is pulled at a constant acceleration of $2 m/s^2$, not with a constant force of $2 N$. If there is no slipping between disks A and B then the rim of B also accelerates at $2 m/s^2$. The angular acceleration of disk B is $\alpha=\frac{2}{2R}=\frac{1}{R} rad/s^2$.

The torque accelerating disk B is $2Rf$ where $f$ is the friction force between the disks. The equation of motion for disk B is therefore
$2Rf=I\alpha=\frac12 M (2R)^2 \alpha=\frac12 \times 2 \times 4R^2 \times \frac{1}{R}$
$2f=4$
$f=2 N$.

answered Dec 20, 2016 by (28,466 points)
edited Feb 18
Right, friction will be 2 N

acceleration of both the disc must be same since there is pure rooling
and friction on smaller disc will be downward and on the bigger dics it will be upwards(Newton's 3rd law) and both will be equal in magnitude.
therefore, writting the equation for bigger dics, f(2R)=2R²/2*(alpha)

                                                                                                  =>, f=2N


since the smaller dics has friction (f) as well as a external force due to which it is accelerating with 2m/s² and the external force was unknown initially. But now it also be found by writing the equation for smaller dics and putting f=2N.

                         "sorry my previous answer was wrong"

answered Feb 1 by (160 points)
reshown Feb 3
Sammy Gerbil's answer is correct. On disk R you get:

$$(F - f)R = I \alpha = \frac{1}{2} M (R)^2 \frac{a}{R}$$

*Note there are two torques acting on disk R*