Answered: A rotary engine is running at 8000…

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter1: Units And Measurement

Section: Chapter Questions

Problem 81AP: Consider the equation s=s0+v0t+a0t2/2+j0t3/6+s0t4/24+ct5/120 , were s is a length and t is a time....

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The topic is all about the Angular Motion of Particles. Please use pen and paper to solve and also use three decimal places. Please show the conversion solution. Thank you.

Here are the formulas and discussion:

θ=θ_o+ω_o t+1/2 αt^2

ω=ω_0+αt

ω^2=ω_o^2+2α(θ-θ_o)

considering the tangential and normal components of the acceleration in a curvilinear motion of particles:

a_t=rα

a_n=rω^2=vω

The angular motion may be derived from the translational motion of particles by considering the given circle:

s=rθ

ds/dt=r dθ/dt or v=rω

dv/dt=r dω/dt or a=rα

P.S. I provide also the answer but show me the complete solution on how to get that answer

A rotary engine is running at 8000 rev/min when it runs out of fuel, causing the engine to decelerate
at a constant rate. If the angular speed is 4000 rev/min after 3200 revolutions, determine the total
time it takes for the engine to stop.

dw
dw do
dw
||
dt
d0 dt
do
..w dw
a do
að + C1 (a is constant)
2
Initial condition: w = 6000 rev/min when 0 = 0. ... C1 = 6000²/2 (rev/min)
1
(교-
6000²) = a0
When 0
= 3600 rev, w
1200 rev/min.
1
(12002 – 6000²) = a(3600)
a = -4800 rev/min?
dw
a =
dt
a dt = dw
at = w + C2
Initial condition: w = 6000 rev/min when t = 0. :. C2 = -6000 rev/min
%3D
When w = 0:
at = C2
- 4800t
-6000
t = 1.250 min
75.0 s
%3D

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