Mechanical Engineering - Theory of machines - Discussion
Discussion Forum : Theory of machines - Section 6 (Q.No. 2)
2.
If the rotating mass of a rim type flywheel is distributed on another rim type flywheel whose mean radius is half the mean radius of the former, then energy stored in the latter at the same speed will be
Discussion:
4 comments Page 1 of 1.
Sidhu said:
6 years ago
Thanks @Chandu.
Sadashiv said:
6 years ago
Thanks for explaining @Chandu.
Rupesh said:
7 years ago
Thank you @Chandu.
Chandu said:
1 decade ago
First flywheel mean radius = x.
Second flywheel mean radius = x/2 (condition given).
Energy storage (ES) = (1/2)I*omega^2.
ω1 = ω2 (condition given).
Energy storage first flywheel mean (ES) = I = mr^2 = mx^2.
Energy storage second flywheel mean (ES) = I = mr^2 = mx^2/4.
So 1/4th of the first one.
Second flywheel mean radius = x/2 (condition given).
Energy storage (ES) = (1/2)I*omega^2.
ω1 = ω2 (condition given).
Energy storage first flywheel mean (ES) = I = mr^2 = mx^2.
Energy storage second flywheel mean (ES) = I = mr^2 = mx^2/4.
So 1/4th of the first one.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers