# Rotational Dynamics of a grinding wheel

1. Mar 31, 2009

### 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?

(ii) At what time did the wheel stop?

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

2. Mar 31, 2009

### rl.bhat

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

3. Mar 31, 2009

### 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

(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.

4. Mar 31, 2009

### rl.bhat

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

5. Mar 31, 2009

### 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.

6. Mar 31, 2009

### 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.

7. Mar 31, 2009

### 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.

8. Mar 31, 2009

### 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.

9. Mar 31, 2009

### Black_Hole???

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

10. Mar 31, 2009

### rl.bhat

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

11. Mar 31, 2009

### 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

12. Mar 31, 2009

### Black_Hole???

v^2 -u^2 = 2as

What is the u^2?

13. Mar 31, 2009

### rl.bhat

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