Pendulum Motion TBR CBT2 Question

[glam407]

11. Which of the following graphs represents the tension in the cable as a function of displacement angle as a pendulum swings through a complete cycle?

Is apparently the answer, and I sort of understand the explanation. "The tension in the cable varies with the angle of the cable with respect to the surface to which the cable is attached. When the pendulum bob points straight down, the tension must offset the entire weight as well as counterbalance the centripetal force, because there is no net force on the pendulum bob. However, when the bob is at its highest point (and motionless), the tension is reduced, because there is no centripetal force to offset. Only the cosine portion of the weight must be offset, since there is a net force on the pendulum bob. This means that the tension magnitude constantly oscillates between a maximum and minimum. This eliminates choices A and B. Because tension is maximum when the cable points straight down, the graph must show a maximum tension when the displacement angle, θ, is equal to zero. The best answer is D."

But why is there no centipetal force at the top ( or an angle) of a pendulum?

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[Mehd School]

Of the plethora of critical thinking questions they can ask about a pendulum, I'd be very surprised that your MCAT will be interested in Tension here. Perhaps I'm wrong.

 

[Captain Sisko]

11. Which of the following graphs represents the tension in the cable as a function of displacement angle as a pendulum swings through a complete cycle?

Is apparently the answer, and I sort of understand the explanation. "The tension in the cable varies with the angle of the cable with respect to the surface to which the cable is attached. When the pendulum bob points straight down, the tension must offset the entire weight as well as counterbalance the centripetal force, because there is no net force on the pendulum bob. However, when the bob is at its highest point (and motionless), the tension is reduced, because there is no centripetal force to offset. Only the cosine portion of the weight must be offset, since there is a net force on the pendulum bob. This means that the tension magnitude constantly oscillates between a maximum and minimum. This eliminates choices A and B. Because tension is maximum when the cable points straight down, the graph must show a maximum tension when the displacement angle, θ, is equal to zero. The best answer is D."

But why is there no centipetal force at the top ( or an angle) of a pendulum?

at the top of the arc the velocity is zero. angular acceleration and thus centripetal force are zero. a pendulum is a great way to combine concepts relating to periodic motion and mechanics, so you could very well see something like this. good luck!

 

[BerkReviewTeach]

Of the plethora of critical thinking questions they can ask about a pendulum, I'd be very surprised that your MCAT will be interested in Tension here. Perhaps I'm wrong.

You never know what they'll ask, until of course you've taken the test.

The test writers specialize is asking conceptual questions about all aspects of a topic. Their questions can address a very basic idea about a system and ask it in an atypical or typical way. A question on the tension in a cable within a pendulum seems very fair and probable, much like a question on any other aspect of a pendulum system seems fair. If you start guessing what aspects of a concept are more apt to be tested, you run the risk of missing vital points of that material.

 

[Postictal Raiden]

at the top of the arc the velocity is zero. angular acceleration and thus centripetal force are zero. a pendulum is a great way to combine concepts relating to periodic motion and mechanics, so you could very well see something like this. good luck!

I thought that the velocity would be maximum when the pendulum is vertical (zero angle of displacement) and would be zero at the maximum angle of displacement. From what I know, KE is maximum at zero displacement and PE is maximum at max displacement.

To answer the question:

F(centripital) = Tension - mg cos
Tension = Fc + mg cos
From the above equation, we can see that when the angle of displacement is zero mg will be maximized and tension will be at minimum. In other words, Tension is highest when the angle is zero and lowest when the angle is at maximum.

 

[mehc012]

I thought that the velocity would be maximum when the pendulum is vertical (zero angle of displacement) and would be zero at the maximum angle of displacement. From what I know, KE is maximum at zero displacement and PE is maximum at max displacement.

To answer the question:

F(centripital) = Tension - mg cos
Tension = Fc + mg cos
From the above equation, we can see that when the angle of displacement is zero mg will be maximized and tension will be at minimum. In other words, Tension is highest when the angle is zero and lowest when the angle is at maximum.

You guys were saying the same thing about velocity..."top of the arc" = "maximum angle", as the angle = 0 at the lowest point of the swing.

I am confused by the bolded statement, however, since it seems contradictory both to your equation and to your next sentence. Either my reading comprehension sucks (very possible, it's my biggest weakness on word problems) or something got mistranslated there!

To restate what everyone else has said, however, there is no Fc at the highest point because v=0:

Fc = ma
a=v²/r
v=0

Plug 'n chug and you get Fc = 0

 

[Captain Sisko]

You guys were saying the same thing about velocity..."top of the arc" = "maximum angle", as the angle = 0 at the lowest point of the swing.

I am confused by the bolded statement, however, since it seems contradictory both to your equation and to your next sentence. Either my reading comprehension sucks (very possible, it's my biggest weakness on word problems) or something got mistranslated there!

To restate what everyone else has said, however, there is no Fc at the highest point because v=0:

Fc = ma
a=v²/r
v=0

Plug 'n chug and you get Fc = 0

you got it.

 

[SN2ed]

Thread closed. These kinds of questions belong in the Study Q&A forum only.