Question

During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 32.932.9 rad/s. Find the angular displacement of the tub during a spin of 60.960.9 s, expressed both in radians and in revolutions.

Answer #1

ωi =32.9329 rad/s

steady angular velocity so α = 0

t= 60.960 sec

θ =ωi*t +1/2*α*t^{2} = ωi*t +0 =ωi*t = 32.9329* 60.960 =
**2007.58958rad = 2008 rad**

one revolution = 2pi rad

so

number of revolutions =**2007.58958 /(2pi) =319.517805 =
320 revolutions**

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Goodluck Comment in case any doubt, will reply for
sure..**

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The tub of a washer is in a spin-dry cycle, running at a
constant angular speed of 50 rad/s for a duration of 10s. After
10s, the lid is opened and the safety switch turns off the washer.
The tub slows down to rest in 5s. Find the total angular
displacement (for entire 15s) in (a) radians and (b)
revolutions?
SHOW ALL WORK ON HOW YOU GOT ANSWER PLEASE

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interval?

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and stopping, determine the total number of revolutions undergone
by the tub during this entire time interval.

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switch turns off the washer. The tub slows to rest in 11.0 s.
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interval? Assume constant angular acceleration while it is starting
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rest and reaching an angular speed of 4.0 rev/s in 13.0 s. At this
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switch turns off the washer. The tub slows to rest in 14.0 s.
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PRACTICE IT
A wheel rotates with a constant angular acceleration of 3.25
rad/s2. Assume the angular speed of the wheel is 2.20
rad/s at ti = 0.
(a) Through what angle does the wheel rotate between t
= 0 and t = 2.00 s? Give your answer in radians and
revolutions.
___________ rad
___________ rev
(b) What is the angular speed of the wheel at t = 2.00
s?
___________rad/s
(c) What angular displacement (in revolutions) results while the
angular...

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