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TPR Physics Practice problem help.


Airjordan

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Hey, I was wondering if anyone with the 2011 review book for physics could help me out on a question. It's at the end of chapter 4 in the practice passage question #4, I don't understand how the radius they got is 20 or w.e they're multiplying by 20 is in the solution :S Thanks.

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  • 1 year later...

I'm having the same problem with this question. I'm extremely frustrated with myself because I'm not seeing how the answer provided in the book found r to be 20.

 

The related passage material is:

A ride consists of a carriage with mass 300 kg and with maximum occupancy of 300 kg. The carriage is attached to a mechanical arm of length L=5 m that is capable of rotation. The arm is able to provide the torque necessary to swing the riders back and forth on a circular path. Initially, the trips back and forth are very small, but with each trip the swings become around larger. Eventually, the riders have enough momentum to swing 360 degrees around, performing a complete circle. In order to partake in this ride, the passengers must be restrained in their seat.

 

The question:

With a full carriage, the ride suffers a power outage with the mechanical arm perpendicular to the horizontal. How much torque must the mechanical arm provide in order to prevent the passengers from swinging down? (Assume the mechanical arm itself does not require any torque support).

 

I get that in order to keep itself from swinging down, the torque needs to cancel out the torque caused by gravity.

I also get that it'd be: torque = rFsin(theta) = mgr = 600 X 10 X r....here is my problem. I thought that r is going to be 5 m...but the book says r is 20 m. Any help would be much appreciated =)

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  • 1 month later...

Was doing this question and decided to google it out of frustration. Found this topic.

 

I think the 20 is a mistake. Should be 4.

 

Here's a pic of the passage/diagram (Cool&Spicy typed the question out in the third post of this topic), for anyone wondering:

 

2vb38jn.jpg

 

What I don't understand is how gravity provides a torque.

 

If the arm is perpendicular to the horizontal, isn't feta = 180, so sin 180 = 0?

 

So torque = rFsinfeta = 0? Like medschool_2014 pointed out.

 

I'm missing something obvious here.

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Was doing this question and decided to google it out of frustration. Found this topic.

 

I think the 20 is a mistake. Should be 4.

 

Here's a pic of the passage/diagram (Cool&Spicy typed the question out in the third post of this topic)' date=' for anyone wondering:

 

 

What I don't understand is how gravity provides a torque.

 

If the arm is perpendicular to the horizontal, isn't feta = 180, so sin 180 = 0?

 

So torque = rFsinfeta = 0? Like medschool_2014 pointed out.

 

I'm missing something obvious here.[/quote']

 

No, you're not missing anything. This is actually a known error, and you're correct for both reasons.

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I dont have the book but if you posted up the question i might be able to help. It'll prob be easiest to put up a picture rather than typing it outKl4IhB

j8cu

 

Basically it was a question referring to Figure 2, and the answer used r=20m for some reason

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