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I am currently taking a basic physics course in college and I am having a bit of trouble on this problem that deals with rotational and translational kinetic energy. Let's begin:

The question: Two disks problem

The Answer:

They both have the same amount of kinetic energy. (Letter B)

Where I am having trouble:

Alright, I understand that that $\text{Work} = \Delta\mathrm{KE}$, where $\Delta\mathrm{KE}$ is the change in kinetic energy from start to end. I also know that $\text{Work} = \vec F \cdot \vec D = ‖F‖\, ‖d‖\sin\theta$, where $\vec F$ and $\vec D$ are the Force and distance vector respectively.

As we can see from the picture, both disks have the same force being applied to them and they also go the same distance $\vec d$. Now because the force vector and distance vector are parallel to each other, finding work (and subsequently $\Delta\mathrm{KE}$) is simply the magnitude of the force vector multiplied by the magnitude of the distance vector.

So, this shows us that both disks have the same amount of kinetic energy (since they start from rest, both of their total kinetic energies are equal). But I have a problem with this:

Disk 1 only has translational kinetic energy, where as Disk 2 has rotational and translational kinetic energy.

\begin{align} \mathrm{KE}_\mathrm{disk 1}&=\frac{1}{2}mv^2\\ \mathrm{KE}_\mathrm{disk 2}&=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 \end{align}

Now because we found that both disks have the same amount of kinetic energy (and the same mass), that means that they have the same translational speed. (In fact, my professor also did a demo of this in class and we observed that they did have the same speed).

My problem is, is that Disk 2 also has rotational kinetic energy and because Total Kinetic Energy (Total KE) is the summation of the translational and rotational KE, why doesn't Disk 2 have more energy than Disk 1?

Since I know they have the same Total KE, then I am guessing Disk 2 doesn't have any rotational KE, but how is this possible?

Where is my logic falling apart here? What assumptions did I make incorrectly and can someone please correct my understanding?

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  • $\begingroup$ How does the rope connect to the disk 2? $\endgroup$ – lucas Apr 20 '16 at 15:22
  • $\begingroup$ If the rope is connected to disk 2 like disk 1, then disk 2 after pi/2 rotating will moves like disk 1 and doesn't rotate. $\endgroup$ – lucas Apr 20 '16 at 15:25
  • $\begingroup$ I stopped reading when you said this: "both disks have the same force being applied to them and they also go the same distance d." That is not what the problem says! $\endgroup$ – garyp Apr 20 '16 at 16:54
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While you are correct saying disk 2 has rotational kinetic energy, you are missing that no matter the situation, since ground is frictionless, the work done be external force(F, in this case), is same in both cases. Thus by work energy theorem, $$Work=change in KE$$.

Thus since work in case 1 equals work in case 2 thus both disk have same kinetic energies.

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"As we can see from the picture, both disks have the same force being applied to them and they also go the same distance d→"

This is the erroneous assumption - the 2 discs do not go the same distance. Some of the distance that the rope is pulled will rotate disc 2 as it unravels. As a result the liner distance is less and the balance of work goes into the rotational acceleration.

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When connected to the centre of mass of disc 1 the force causes an acceleration of the centre of mass and the work done by the force is $Fd$ where $d$ is the displacement of the centre of mass and the force $F$. The translational kinetic energy of the disc increases by an amount $Fd$.

When the force acts on the rim of disc 2 the centre of mass of the disc undergoes a translational movement of $x$ and rotates through an angle $\theta$. If the radius of the disc is $R$ then the total movement of the force is $x+ R\theta = d$. The total amount which has been done the force is the same but work $Fx$ has increased the translation kinetic energy of the disc and work $FR\theta$ has increased the rotational kinetic energy of the disc.

So the work done by the force is the same in both cases but all of work $Fd$ has increased the translation kinetic energy of disc 1 and only some of work $Fd$ has increased the translational kinetic energy od disc 2, the remained of the work increasing the rotational kinetic energy of disc 2.

Since the work done on both discs is the same $=Fd$ the total kinetic energy of both discs is also the same.

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@Prayas Agrawal

No I'm not missing that - I'm just explaining the original error. Of course the total work = the KE.

In fact since the floor is frictionless, the disc remains stationary and just spins since the force has no line of action through the CoM. In the case of friction the force of the weight of the disc provides a torque on the ground via the coeff of friction up to the limit of mg.mu (mu = coeff of friction), then the disc slips. With mu = 0 the disc just slips.

Also - your point is spurious since friction in this case would not be dissipative unless F > mu.mg (i.e. sliding). Before that the friction force just provides a fulcrum for the moment with no loss of energy.

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  • $\begingroup$ "you are missing" was not informed to you, but was informed to the asker. $\endgroup$ – Prayas Agrawal Apr 24 '16 at 17:05
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I think the error occurs where you state :

Now because we found that both disks have the same amount of kinetic energy (and the same mass), that means that they have the same translational speed. (In fact, my professor also did a demo of this in class and we observed that they did have the same speed).

The two discs have the same total KE but not the same translational KE so the translational speeds are also different. If the string unwinds from Disk 2 and the end of the string moves the same distance as for Disk 1, then the CM of Disk 2 cannot move the same distance as the CM of Disk 1, so the translational speed cannot be the same.

What kind of 'demo' did your professor do? If it used software to simulate the situation, the result depends on how the problem was set up. It seems to me that the 'demo' must have been faulty.

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