Tension question (this should be basic)

  • Thread starter Thread starter pureouchies4717
  • Start date Start date
  • Tags Tags
    Tension
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving tension in a rotating system. A steel block is attached to a hollow tube and experiences a thrust force while rotating on a table. Participants are tasked with determining how many revolutions the block makes before the tube breaks, given specific parameters such as mass, thrust force, and maximum tension.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to calculate angular acceleration and forces acting on the block, including friction and tension. Some express confusion over the relationships between variables and how to derive the number of revolutions from their calculations.

Discussion Status

Several participants have shared their calculations and approaches, but there is no clear consensus on the correct method or solution. Some have provided equations related to angular motion and forces, while others express uncertainty about the concepts involved, such as moment of inertia and the meaning of specific terms.

Contextual Notes

Participants are working under time constraints, with one expressing urgency due to a limited timeframe for assistance. There is also mention of varying levels of familiarity with the underlying physics concepts, such as inertia.

pureouchies4717
Messages
98
Reaction score
0
tension question on object moving in a circle (this should be basic)

A .6 kg steel block rotates on a steel table while attached to a 1.20 m-long hollow tube. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.10 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 50.0 N. (coefficient of kinetic friction:.6) If the block starts from rest, how many revolutions does it make before the tube breaks?

knight_Figure_07_55.jpg


im stumped...

look 1 post down... I am trying
 
Last edited:
Physics news on Phys.org
ah i found something:

at= 6.833333 m/s^2

Fr= mat - Ff

Fr= 4.1N- Ff
= 4.1-3.528
=.572N

a(c)= .953333m/s^2
 
Last edited:
ok here's what i tried:

at 50N, the object is going at:
v= 10m/s

so i plugged this in the T=(mv^2)/r equation and got: T=50s, but this looks wrong, and i don't know how to get revs from this...
 
hmmm...

angluar velocity= (at/r)t= 284.722rad/s= 2718.89 rpm

2718.89= 45.315 revs/second

45.315 x 50 = 2265.75 revolutions (wrong)
 
Last edited:
guys please help me, there's only one hour left
 
Here's a start:

The angular acceleration is given by

[tex]I \frac {d^2 \theta}{dt^2} = r (F_T - \mu mg)[/tex]

where I is the moment of inertia (essentially just [itex]m R^2[/itex]). Use the equation solve for the angular velocity and the angular displacement as a function of time then determine the time at which the centripetal force reaches the breaking point.
 
Tide said:
Here's a start:

The angular acceleration is given by

[tex]I \frac {d^2 \theta}{dt} = r (F_T - \mu mg)[/tex]

where I is the moment of inertia (essentially just [itex]m R^2[/itex]). Use the equation solve for the angular velocity and the angular displacement as a function of time then determine the time at which the centripetal force reaches the breaking point.


thank you , thank you for responding
 
Tide said:
Here's a start:

The angular acceleration is given by

[tex]I \frac {d^2 \theta}{dt^2} = r (F_T - \mu mg)[/tex]

where I is the moment of inertia (essentially just [itex]m R^2[/itex]). Use the equation solve for the angular velocity and the angular displacement as a function of time then determine the time at which the centripetal force reaches the breaking point.

thanks so much, but i don't really get what youre saying. what does Ft mean? o, i didnt learn about inertia yet
 
Last edited:
guys, i don't want you to waste your time. if you arent able to help in the next 30 minutes, don't even worry about it
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 11 ·
Replies
11
Views
7K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 1 ·
Replies
1
Views
7K
Replies
1
Views
7K
  • · Replies 11 ·
Replies
11
Views
11K