# Pure rolling of a cylinder on a rough surface.

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A solid cylinder of mass $6 kg$ lies on a rough horizontal surface. The coefficient of friction $\mu = 0.6$. A constant force acts horizontally on the cylinder. The line of action of the force $F$ is at height $\dfrac{2}{3}R$ above the centre of the cylinder . Find the maximum value of $F$ if the cylinder rolls without slipping.

Attempt:

According to me friction should act in the forward direction because $k^2\ < Rx$, where $k$ is radius of gyration and $x$ is perpendicular distance of line of action of force from centre.

$\implies F + f= ma \implies F= ma -f$ , thus the value of $f$ is maximised when friction acting is $0$.

I also have the torque equation:

$\tau _{net}= I\alpha \implies F\dfrac{2}{3}R - fR= \dfrac{1}{2}MR^2 \alpha$ where $\alpha = \dfrac{a}{R}$ for pure rolling.

But substituting $f=0$ led to a contradiction.

How should I solve this problem then?

asked Nov 3, 2017
edited Nov 3, 2017

## 1 Answer

1 vote

Your assumption that $F$ is maximum when $f=0$ is incorrect. $F$ is maximised for a fixed value of $a$ when $f=0$, but there is nothing in the question which tells us that $a$ has a fixed value.

Using the 3 equations you have got, eliminate $a$ and $\alpha$ to get an expression for $F$ in terms of $f$. Then note that $f\le \mu Mg$. You will get an inequality for $F$ in terms of $\mu Mg$.

answered Nov 3, 2017 by (28,448 points)
edited Nov 3, 2017