Giancoli's Physics: Principles with Applications, 6th Edition

8

Rotational Motion

Change chapter8-1: Angular Quantities

8-2 and 8-3: Constant Angular Acceleration; Rolling

8-4: Torque

8-5 and 8-6: Rotational Dynamics

8-7: Rotational Kinetic Energy

8-8: Angular Momentum

8-9: Angular Quantities as Vectors

Problem 34

A

a) $2.10 \times 10^{-3} \textrm{kgm}^2$

b) $0.0718 \textrm{Nm}$

b) $0.0718 \textrm{Nm}$

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COMMENTS

By Dfirst05 on Fri, 2/13/2015 - 4:57 AM

On part B shouldn't the final answer be Applied Torque = 0.0598 Nm since you are adding a negative frictional torque?

By Mr. Dychko on Sat, 2/14/2015 - 12:13 AM

Hi Dfirst05, thanks for the question about negative signs with torque. Depending on your perspective, the video has a slight error. Let me explain. At 3:59, I mention that the frictional torque $\tau_{fr} = 0.005983 \textrm{ N m}$, positive. Technically, you are correct in saying the frictional torque should be negative since I took the angular acceleration due to friction to be negative. As a matter of personal preference, and as an obvious source of confusion for students in this case (sorry!), I just wanted the *magnitude* of the frictional torque, without caring about it's sign. The reason for this is that later, when I state the *net torque*, I wrote it as $\tau_{applied} - \tau_{fr} = I \alpha$, and take notice of the **minus** in this equation. That minus in the equation accounts for the direction. My personal preference is to write Net Force or Net Torque with pluses and minus **in the equation** to account for direction, and then always substitute **magnitudes** in for the values. Not everyone does it this way. The alternative is to write the net torque as $\tau_{applied} + \tau_{fr} = I \alpha$ and substitute a negative frictional torque. So the two ways to a correct answer are:

This is the option used in the video. In this case $\tau_{applied} - \tau_{fr} = I \alpha$ and torques are always positive, no matter what, since their direction are accounted for in the equation.

This option is more common in textbooks. In this case $\tau_{applied} + \tau_{fr} = I \alpha$ and the frictional torque would be substituted as a negative number. With this approach you would be **subtracting** a negative frictional torque at 6:22 in the video since the final equation would be $\tau_{applied} = I \alpha - \tau_{fr}$

With either approach, the correct answer is $0.0718 \textrm{ N m}$. Both approaches for doing net torque (or net force) are common, and it's just important to recognize which approach is being used, and being consistent.

I hope this helps,

Mr. Dychko

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