A wheel is making 240 rpm and after 30 sec and after 30 sec

[manal950]

1 ) A wheel is making 240 rpm and after 30 sec and after 30 sec it is running at 150 rpm .
A) Find the time elapse before it stops if the retardation is uniform ?
B) Find the time elapse after it stops if the retardation is uniform ?

my answer :

w0 = 240 rpm = 8Pi rad/s
w = 240 rpm = 5pi rad/s
t = 30

------

w = w0 + at
a = -0.1Pi

time elapse before it stops
w = w0 + at
5pi = 8Pi - 0.1Pi (t)
t = 94.4 s

time elapse after it stops
w = w0 + at
5pi = 0 - 0.1Pi (t)
t = 50 s

 

[grzz]

correct

 

[Curious3141]

1 ) A wheel is making 240 rpm and after 30 sec and after 30 sec it is running at 150 rpm .
A) Find the time elapse before it stops if the retardation is uniform ?
B) Find the time elapse after it stops if the retardation is uniform ?

Something is wrong, or missing, from the question. Why is "and after 30 sec" mentioned twice? Also, part A makes sense, but part B does not. After the wheel stops, it remains stationary and that's it - what elapsed time are they expecting one to calculate?

my answer :

w0 = 240 rpm = 8Pi rad/s
w = 240 rpm = 5pi rad/s
t = 30

------

w = w0 + at
a = -0.1Pi

This is correct so far. I presume you did this:

8\pi = 5\pi + a(30)

to get this value of a?

time elapse before it stops
w = w0 + at
5pi = 8Pi - 0.1Pi (t)
t = 94.4 s

How can this be right? You're basically using the same equation as above, but you're getting a different value for t! You should just get t = 30s, but you're just unnecessarily recalculating a value already given in the question.

time elapse after it stops
w = w0 + at
5pi = 0 - 0.1Pi (t)
t = 50 s

This is correct. Specifically, this is the additional time the wheel takes to come to a rest after it's already slowed to 150 rpm. The total time taken to slow down from 240 rpm to 0 rpm is 80s, but that can just be calculated by adding 30s and 50s.

The much simpler way to do this is to simply say that if the wheel takes 30s to retard from 240 rpm to 150 rpm, the retardation is \frac{(240-150) rpm}{30s}= 3 rpm/s.

Hence to come to a complete stop, the wheel would take: \frac{(240-0) rpm}{3 rpm/s} = 80s.

 

[manal950]

1 ) A wheel is making 240 rpm and after 30 sec it is running at 150 rpm .
A) Find the time elapse before it stops if the retardation is uniform ?
B) Find the time elapse after it stops if the retardation is uniform ?

sorry now the question correct

what is you mean by but part B does no has sense "

 

[manal950]

thaaaaaaaaaanks so much

 

[Curious3141]

1 ) A wheel is making 240 rpm and after 30 sec it is running at 150 rpm .
A) Find the time elapse before it stops if the retardation is uniform ?
B) Find the time elapse after it stops if the retardation is uniform ?

sorry now the question correct

what is you mean by but part B does no has sense "

t = 0s, wheel is spinning at 240 rpm
t = 30s, wheel has slowed to 150 rpm
t = 80s, wheel has slowed to a complete stop
t = 81s, wheel remains stopped
t = 90s, wheel remains stopped
t = 100s, wheel remains stopped
...
t = ∞ , wheel remains stopped.

There's nothing new happening after the wheel comes to a stop, so what is part B) expecting?

It is entirely possible that the wheel after instantaneously coming to a stop, is immediately made to spin in the other direction, but this is not specified in the question, and one cannot assume it. And even then, the question still doesn't make sense.

Do you understand what part B) is asking for? If so, please explain it to me. But if, like me, you also don't understand, please clarify with your instructor/teacher.