Torque, gravity, and equilibrium

AI Thread Summary
The discussion revolves around calculating the torque required to prevent passengers from swinging down when a ride's mechanical arm is perpendicular to the horizontal during a power outage. The initial calculations suggest using a lever arm of 5 meters, leading to a torque of 30,000 N·m, while the book provides a solution of 120,000 N·m, possibly using a radius of 20 meters. Participants express confusion over the problem's wording, particularly regarding the arm's orientation and the correct interpretation of the scenario. It is noted that the problem may be poorly worded, leading to misunderstandings about the mechanics involved. Overall, the conversation highlights discrepancies in problem interpretation and the importance of clarifying physics concepts.
musicalmedicin
Messages
2
Reaction score
0

Homework Statement


Description:

A ride consists of a carriage with a mass 300kg and with maximum occupancy of 300kg. The carriage is attached to a mechanical arm of length L = 5m 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 larger. Eventually, the riders have enough momentum to swing 360° around, performing a complete circle. In order to partake in this ride, the passengers must be restrained to their seats.

Problem statement:

With a full carriage, the ride suffers a power outage withe 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.)

Homework Equations



τ = rFsinθ

τ = ℓF

Fgrav = mg

ac=V2/r

The Attempt at a Solution


I first started by determining that the only force the arm would have to oppose was gravity. Since the arm is currently perpendicular to the horizontal We can use the lever arm equation, τ = ℓF and set it equal to Fgrav
∴τ = ℓmg = (5m)(300kg + 300kg)(10m/s2) = 3 x 104 N·m

The book I am using to study provides answer in the solution section of 12 x 10^4. It also shows how this answer was achieved. And shows τ = rFsinθ = mgr = (300 + 300) x 10 x 20 = 12 x 104 N·m. My interpretation of this answer is that the book's author is using 20 meters as the radius while I am using the 5 meters that is described in the description. I have been unable to find any information earlier in the chapter that makes me believe that I should be using 20 meters. Can someone please point out any flaws in my logic and calculations that would lead me to improper use 5 meters instead of 20 meters?

Thank you!
 
Last edited:
Physics news on Phys.org
I'm not sure what "perpendicular to the horizontal" means. (I would call that vertical!) I assume that they meant that the arm was horizontal. If so, I agree with your answer.
 
Google to the rescue. Apparently it's a known problem with the book. At least that's what it says here...

http://premed101.com/forums/archive/index.php?t-52896.html

PS: I think the problem is very badly worded.
 
Last edited by a moderator:
My reading of the problem is that the power fails as the carriage passes through the low point (eg arm vertically downwards) with some velocity. The arm will continue to rotate using up KE until it comes to a halt at some height/angle. They want to know the max torque that could be required to stop it falling back down from that point.

That will occur (rather obviously) if it reaches the horizontal position before falling back. So yes I agree with your answer.
 
Thank you both for your responses! CWatters, I didn't even think to google the question because Princeton Review has a section of the online access that is suppose to describe any mistakes that have been found throughout the books, since it wasn't in there I assumed I made an error.
 
Good find, CWatters! :cool:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top