{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

136_pdfsam_math 54 differential equation solutions odd

# 136_pdfsam_math 54 differential equation solutions odd -...

This preview shows page 1. Sign up to view the full content.

Chapter 3 17. Since the motor is turned off, its torque is T = 0, and the only torque acting on the ﬂywheel is the retarding one, 5 ω . Then Newton’s second law for rotational motion becomes I dt = 5 ω with ω (0) = ω 0 = 225 (rad / sec) and I = 50 ( kg / m 2 ) . The general solution to this separable equation is ω ( t ) = 5 2 I t + C = 0 . 05 t + C. Using the initial condition, we find ω (0) = 0 . 05 · 0 + C C = ω (0) = 225 = 15 . Thus t = 1 0 . 05 15 ω ( t ) = 20 15 ω ( t ) . At the moment t = t stop when the ﬂywheel stops rotating we have ω ( t stop ) = 0 and so t stop = 20(15 0) = 300 (sec) . 19. There are three forces acting on the object: F 1 , the force due to gravity, F 2 , the air resistance
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online