Finding uniform transmission condition

Click For Summary
SUMMARY

The discussion centers on the analysis of constant mechanical power transfer in linear motion, contrasting it with established rotational gear transmission principles. It concludes that for linear acceleration under constant power, the kinetic energy increases linearly with time, represented mathematically as 0.5mv² = Bt, where B is a constant. The acceleration (a) is determined to be inversely proportional to the square root of time, leading to the conclusion that the force applied is proportional to the square root of mass divided by time.

PREREQUISITES
  • Understanding of classical mechanics principles, particularly kinetic energy.
  • Familiarity with the concept of linear acceleration and its mathematical representation.
  • Knowledge of differential calculus as it applies to motion equations.
  • Basic grasp of power transfer concepts in mechanical systems.
NEXT STEPS
  • Research the implications of constant power transfer in linear systems.
  • Explore advanced topics in differential calculus related to motion equations.
  • Study the differences between linear and rotational dynamics in mechanical systems.
  • Investigate real-world applications of constant mechanical power in engineering designs.
USEFUL FOR

Mechanical engineers, physics students, and anyone interested in the principles of motion and energy transfer in mechanical systems.

vin300
Messages
602
Reaction score
4
We have never discussed about constant mechanical power transfer for the linear case, as against rotational well documented gear transmission. The energy acquired by linear motion, if varies linearly, then 0.5mv^2 must have constant differential. Trying it, if u=0, v=at.
d/dt[0.5m(at)^2]=constant. m(a^2)t=const.Hence concluding, a is inversely proportional to root of time, and the force to be applied is proportional to {sqrrt.(m/t)}. Does this look fine?
 
Physics news on Phys.org
Who is "we"?

"The energy acquired by linear motion, if varies linearly, then 0.5mv^2 must have constant differential "
My interpretation: if you apply a constant power for linear acceleration, then the kinetic energy increases linearly with time.

v = at doesn't fly, though: that's only for constant acceleration. Here if ##{1\over 2} mv^2 = B\, t## with ##B## constant.
However, ##a## is still ##\propto \sqrt{1/t}##, but not ##\propto \sqrt{m}##.

Do check my claims, please !

--
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
Replies
3
Views
1K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
17K
  • · Replies 32 ·
2
Replies
32
Views
8K