Rotational Dynamics of a grinding wheel

[Black_Hole???]

Ok i am not good with physics and these are some review problems for my test coming up. Please help.

At time t1 = 0 an electric grinding wheel has an angular velocity of 28.0 rad/s. It has a
constant angular acceleration of 32.0 rad/s2. At time t2 = 2.20 s a circuit breaker trips, and the
wheel then turns through another 350 rad as it coasts to a stop at constant angular acceleration.

(i) Through what total angle did the wheel turn between t = 0 and the time it again
stopped?




ANSWER: _______________ Rad

(ii) At what time did the wheel stop?





ANSWER: ________________ ____

(iii) What was the angular acceleration of the wheel as it slowed down?






ANSWER: ________________ ____


answers are A1) 489 rad A2) 9.31 sec A3) -13.83 rad/sec^2
Please show work.

 

[rl.bhat]

Please show work.
First of all please show your attempts. What are relevant equations in rotational motion can be applied in this problem?

 

[Black_Hole???]

Q1(i) Through what total angle did the wheel turn between t = 0 and the time it again
stopped?

My attempt was:

28.0 + 32.0(2.20)^2= 182.88 + 350 = 532.88 rad

The answer was suppose to be 489 Rad.

(ii) At what time did the wheel stop?

My attempt was

2.20 + 350/? = ?

I am not sure how i am suppose to find the decrease in acceleration of the angular velocity because i don;t know how long it took and its not given.


(iii) What was the angular acceleration of the wheel as it slowed down?

I can't solve for this until i know answer B.

 

[rl.bhat]

28.0 + 32.0(2.20)^2
For this which formula you are using?

 

[Black_Hole???]

Well for that i was using the

Change in distance = initial velocity + 1/2 a t^2

Then i added that to the 350 Rad it said it traveled once the breaker was tripped.

But i realized i forgot the 1/2 in my equation and end up getting 450.44. Which ends up being wrong again.

 

[rl.bhat]

Change in distance = initial velocity + 1/2 a t^2
It should be
Change in distance = initial velocity*t + 1/2 a t^2
Also find the velocity at 2.2 s. This is necessary for the second part.

 

[Black_Hole???]

Ok The velocity at 2.2 sec is =732 rad/s^2

I got that by

Vf=V0 + at
Vf = 28.0 + (32.0)(2.20)

However i realized that I couldn't find an equation that didn't have both time and acceleration involved. What would i do to isolate just time while using the Initial velocity as 732 rad/sec^2 and the final velocity as 0 rad/sec^2. With the total distance traveled to be 350 rads?

It doesn't give me a mass or a frictional force to work with.

 

[rl.bhat]

Ok The velocity at 2.2 sec is =732 rad/s^2
This is wrong. Is it the velocity or change in distance or some thing else?

Find out three equations of motion.

 

[Black_Hole???]

wow totally my bad. The answer was 98.4. entered the equation as 2.80+320(2.2)

 

[rl.bhat]

Now you can use v^2 -u^2 = 2as to find a and t.

 

[Black_Hole???]

OK i got the answers for both B and C. I had the right idea but i keep messing up with entering it into my calculator so i need to check that before i post stuff.

However i got the answer to C first then B.

c) Vf^2 = V0^2 + 2 a s

0 = (98.4)^2 + 2(a)(350)
a=-13.8

B) t = (Vf-V0)/a

t= (98.4-0)/13.8
t= 7.13

T(total) = 2.2 + 7.13 = 9.33



If i were suppose to get it in the order of the test goes how would i go about that.

Thank you for your patience with me and help

 

[Black_Hole???]

v^2 -u^2 = 2as

What is the u^2?

 

[rl.bhat]

Initial velocity i.e vo.
You can't find the answers in the given order.