The Physics Philes, lesson 53: Even More on Rotational Dynamics

I’m back, once again, just in time to serve you your weekly dose of physics. As promised last week, I’ll walk you through a problem that will illuminate the rotational dynamics of rigid bodies. Well, I’ll try to anyway.

Let’s say we have a cable wrapped around a uniform solid cylinder – 0.120 m in diameter with a mass of 50 kg – that can spin about an axis. You pull the cable with a force of 9.0 N. If we assume the cable unwinds without stretching or slipping, what is the acceleration?

Our target variable is the acceleration of the cable, so we’ll use the equations we learned about the last couple of weeks. We need to find the relationship between the motion of the cable and the motion of the rim of the cylinder. But first things first: let’s draw a diagram.

Screen shot 2013-06-16 at 12.48.32 PM

And now second things second: what is happening? We’ll assume that the cylinder rotates counterclockwise when the cable is pulled, so we’ll take the counterclockwise rotation as positive. The net force on the cylinder is zero because the center of mass is at rest, and the weight (Mg) and the normal force (n) act along lines through the axis of rotation. So the normal force and the weight don’t produce any torque with respect to the axis. The only force we need to worry about is force F.

If you’ll remember the definition of lever arm, you’ll know that force F has a lever arm equal to the radius of the of the cylinder. So the lever arm is 0.060 m (half the diameter of the cylinder), so the equation we’ll use here is τ = FR. We also need to recall another equation we learned several weeks ago: the moment of inertia. Specifically:

Screen shot 2013-06-16 at 1.22.31 PM

This means that we can use the rotational analog of Newton’s second law for a rigid body. In math, that means:

Screen shot 2013-06-16 at 1.28.01 PM

But we need to solve for α. If τ = FR, we get:

Screen shot 2013-06-16 at 1.36.01 PM

So we know that α = 2F/MR. Now we just have to plug in the numbers.

Screen shot 2013-06-16 at 1.45.55 PM

(Remember that we can add radians on as a unit because it’s a dimensionless quantity.)

There you go! We used the relationship between the motion of the cable and the motion of the rim of the cylinder to find the acceleration of the cable. Easy, right?

So far we’ve been dealing with situations in which the axis is stationary. Next week we’ll start a discussion of what we have to do to solve problems in which the the rigid body is rotating around a moving axis.

Featured image credit: Kylesauris

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Mindy is an attorney and Managing Editor of Teen Skepchick. She hates the law and loves stars. You can follow her on Twitter and on Google+.

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