Problem with gears — One geared wheel spins up another

  • Thread starter Thread starter LCSphysicist
  • Start date Start date
  • Tags Tags
    Gears Wheel
Click For Summary
The discussion revolves around the application of Newton's third law to analyze the interaction between two gears, focusing on torque and energy conservation. The equations presented aim to establish a relationship between the torques and angular velocities of the gears, but the user reports discrepancies in the results. The justification for using mechanical conservation principles is highlighted, emphasizing that the work done by one gear's force is countered by the reaction force of the other. Additionally, the concept of treating gear A as a flywheel is mentioned, illustrating its relevance in industrial applications. The kinetic energy of gear A is crucial for maintaining motion in both gears after contact.
LCSphysicist
Messages
644
Reaction score
162
Homework Statement
I am trying to solve a problem by torques and conversation of energy, but the answers don't match.
Relevant Equations
t = rf
E = constant
1590165238935.png

By third Newton's law, i would imagine that fa = -fb, so
and assuming f to be the average value

torque in A
-ra*f*t = Waf - Wao
torque in b
rb*f*t = Wbf

and

Ia*wo² = Ib*wbf² + Ia*waf²

it Looks right to you? I really appreciate this method that i thought, but the answers don't match, and i don't know what is wrong.

THe justification to use the conservation of mechanic would be: The work done by one force will be equal to the work done by the reaction force, but opposite in signal, so it cancel.
 
Physics news on Phys.org
You could treat gear A as a flywheel.
There are many industrial presses and shear machines that use that principle for conforming and cutting metal.

Please, see:
https://www.engineeringtoolbox.com/flywheel-energy-d_945.html

The kinetic energy of gear A (Joules) will have to keep moving both gears after contact is established.

##E_{kA}=(1/2)(I_A)(\omega_0)^2##
 
Last edited:
  • Like
Likes etotheipi
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
7
Views
2K
  • · Replies 121 ·
5
Replies
121
Views
12K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
18
Views
2K
Replies
2
Views
4K