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--- The friction coefficient =0.1 how many revolutions does the cylinder make until it stops?


So it's obvious that we'll first have to use the relation $I\alpha=\tau (about centre)$
and that the net torques at the contact points will form a couple to deaccelerate the rotation but I'm not able to figure out how to find the normal reaction at the vertical wall's contact point to find its torque...
How do I proceed further?

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  • $\begingroup$ You proceed with a free body diagram as always, and then form the equations of motion. $\endgroup$ – JAlex Dec 22 '20 at 17:25
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Look at a free body diagram, describing all the forces acting on the cylinder.

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In green is the weight, in blue is the normal forces and in pink are the frictional forces.

Now use the equations of motion to balance out the above forces with the inertial forces.

You will notice if the cylinder isn't translating the horizontal acceleration must be zero and thus the friction on the bottom should be equal and opposite of the normal force from the wall.

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The normal force at the wall must equal the friction force at the bottom.

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This is a simple exercise in free-body problems. Since the cylinder is being held by the wall, it is in translational equilibrium. The normal force from the side wall cancels the frictional force from the bottom, whereas the friction along the side wall along with the normal force from the bottom wall balance the cylinder's weight.

These will give you two equations in two unknowns that you can solve for the normal reactions. After that, calculate the torque on the cylinder due to the frictional forces from the side wall and the bottom wall (Hint: they add up) and find the angular retardation of the cylinder by dividing the net torque by the cylinder's moment of inertia about its major axis. Solve a simple angular velocity equation, and you will get an answer,

$$ n_\mathrm{rev} = \frac{\omega_0^2 R}{8 \pi g} \left(\frac{\mu_k + 1/\mu_k}{1 + \mu_k}\right)$$ for the number of revolutions, or something of that sort.

PS: I only did a quick, back-of-the-envelope calculation, so if anybody would like to suggest corrections, I am happy to take them.

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