Pendulum Restoring Force

[StarryNights]

I studied for PS using TPR book, and learned that for a pendulum, it is gravity that provides the restoring force to get it back to the vertical line. So the restoring force would be equal to the horizontal component of gravity = mg sin theta. Therefore acceleration a = g sin theta.

Then I did a Kaplan PS section test, and it asked me for acceleration in terms of theta but set it as a = g tan theta. Their diagram indicates that gravity provides no horizontal force and that the restoring force is actually the horizontal component of tension of the pendulum string = T sin theta. Then they set the mg = T cos theta to balance. So if we do (ma = T sin theta) / (mg = T cos theta) then we get a = g tan theta.

Both ways make sense to me depending on how the diagram is drawn, and the truth is that for smaller angles they come out very close anyway. For example, if I set a as 10 and theta as 30 degrees, then both give answers pretty close to 5. I'm just worried if the MCAT doesn't ask for numbers but if they give answers in the form of variables as above, then which one should I use??

Hopefully this all made sense, and thanks for any help!

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[PiBond]

I studied for PS using TPR book, and learned that for a pendulum, it is gravity that provides the restoring force to get it back to the vertical line. So the restoring force would be equal to the horizontal component of gravity = mg sin theta. Therefore acceleration a = g sin theta.

Then I did a Kaplan PS section test, and it asked me for acceleration in terms of theta but set it as a = g tan theta. Their diagram indicates that gravity provides no horizontal force and that the restoring force is actually the horizontal component of tension of the pendulum string = T sin theta. Then they set the mg = T cos theta to balance. So if we do (ma = T sin theta) / (mg = T cos theta) then we get a = g tan theta.

Both ways make sense to me depending on how the diagram is drawn, and the truth is that for smaller angles they come out very close anyway. For example, if I set a as 10 and theta as 30 degrees, then both give answers pretty close to 5. I'm just worried if the MCAT doesn't ask for numbers but if they give answers in the form of variables as above, then which one should I use??

Hopefully this all made sense, and thanks for any help!

at PEmax (when pendulum is highest in its path) i had always used the restoring force mgsine(theta)...i'm having trouble visualizing the kaplan explanation

 

[StarryNights]

Same, I'm more used to the mg sin theta formula too so sat there scratching my head for a couple of minutes.

 

[godcyning]

I studied for PS using TPR book, and learned that for a pendulum, it is gravity that provides the restoring force to get it back to the vertical line. So the restoring force would be equal to the horizontal component of gravity = mg sin theta. Therefore acceleration a = g sin theta.

Then I did a Kaplan PS section test, and it asked me for acceleration in terms of theta but set it as a = g tan theta. Their diagram indicates that gravity provides no horizontal force and that the restoring force is actually the horizontal component of tension of the pendulum string = T sin theta. Then they set the mg = T cos theta to balance. So if we do (ma = T sin theta) / (mg = T cos theta) then we get a = g tan theta.

When you say mg sin theta, that's the tangential component of gravity. This makes sense to me as what the restoring force should be. It correctly describes the net force on the weight, and is consistent with its acceleration and motion. If you think about it, the restoring force should be maximal at the 3 o'clock position, and there it should be the full force of gravity, mg. If this is at 90 degrees, then yeah, mg sin theta works.

mg tan theta is the horizontal component of tension...except not even, since I don't see how horizontal tension has anything to do with gravity. If you draw the gravity vector, then start at the tip and draw a tension vector with the same vertical component as gravity, then draw a horizontal line to top it off... that horizontal line is mg tan theta.

I don't see the physical meaning of that since you don't know how long the tension vector is. (you could calculate it, but it won't coincidentally have the same vertical component as gravity the way Kaplan seems to be assuming, as I understand it). If those vertical components were equal, the weight would never accelerate vertically.

 

[StarryNights]

at PEmax (when pendulum is highest in its path) i had always used the restoring force mgsine(theta)...i'm having trouble visualizing the kaplan explanation

I couldn't print out the Kaplan explanation to scan so I drew their diagram on paint instead. Sorry the diagram looks like chickenscratch, but I'm just really inept. I don't know if it clarifies anything clearer, but it just shows that they have mg as not providing any restoring force.

The same diagram done by TPR would extend the diagonal path of the pendulum out, and when you connect it with the vertical gravity vector then gravity has a horizontal component that's providing the restoring force.

View attachment Kaplan.docx

 

[IndianVercetti]

The TPR vs. Kaplan diagrams are BOTH correct - it's just a difference in the way the two books define their diagrams. The fact that you can reason through both different diagrams already shows you have an understanding of the topic. The MCAT won't ask a specific question with the assumption that the test-taker MUST draw a particular diagram, as opposed to another equally correct diagram.

What most likely will appear is them asking you to determine the restoring force, based on a given mass and given angle - which you can do with either equation.

 

[Beta Cell]

When you say mg sin theta, that's the tangential component of gravity. This makes sense to me as what the restoring force should be. It correctly describes the net force on the weight, and is consistent with its acceleration and motion. If you think about it, the restoring force should be maximal at the 3 o'clock position, and there it should be the full force of gravity, mg. If this is at 90 degrees, then yeah, mg sin theta works.

mg tan theta is the horizontal component of tension...except not even, since I don't see how horizontal tension has anything to do with gravity. If you draw the gravity vector, then start at the tip and draw a tension vector with the same vertical component as gravity, then draw a horizontal line to top it off... that horizontal line is mg tan theta.

I don't see the physical meaning of that since you don't know how long the tension vector is. (you could calculate it, but it won't coincidentally have the same vertical component as gravity the way Kaplan seems to be assuming, as I understand it). If those vertical components were equal, the weight would never accelerate vertically.

3 and 9 o'clock or 90 and 270 degreees (setting 12 o'clock to be 0).