Angular Acceleration and a grindstone

[bamalam]

HELP! I keep getting a solution that is not part of the choices. Where am I messing up?

A grindstone, originally rotating at 126 rad/s undergoes a constant angular acceleration so that it makes 20.0 rev in the first 8.00s. What is the angular acceleration?


a) 0.313 rad/s2
b) 0.625 rad/s2
c) 2.50 rad/s2
d) 1.97 rad/s2
e) 13.79 rad/s2

My Calculations:
20rev*2PIrad/rev = 40PI rad = theta
t = 8 sec
wo = 126 rad/sec

theta = wo*t+1/2*a*t^2
40PI = 126*8+.5*a*8^2
40PI-126*8 = 32a
a = -27.57 rad/sec^2

 

[Pseudo Statistic]

HELP! I keep getting a solution that is not part of the choices. Where am I messing up?

A grindstone, originally rotating at 126 rad/s undergoes a constant angular acceleration so that it makes 20.0 rev in the first 8.00s. What is the angular acceleration?


a) 0.313 rad/s2
b) 0.625 rad/s2
c) 2.50 rad/s2
d) 1.97 rad/s2
e) 13.79 rad/s2

My Calculations:
20rev*2PIrad/rev = 40PI rad = theta
t = 8 sec
wo = 126 rad/sec

theta = wo*t+1/2*a*t^2
40PI = 126*8+.5*a*8^2
40PI-126*8 = 32a
a = -27.57 rad/sec^2

Let's back up:

A grindstone, originally rotating at 126 rad/s undergoes a constant angular acceleration so that it makes 20.0 rev in the first 8.00s. What is the angular acceleration?

We have u = 126rads^-1
We also have s = 20rev = 40Pi Radians
t = 8s
Now we're left with:
40Pi = (126)(8) + 1/2a(64)
40Pi = 1008 + 32a
a = -27.57300918...rad s^-2

Same answer here...
Perhaps mistake in the question? (Maybe..)

 

[quasar987]

I have -27.57 too. At any rate, none of the proposed answer have a minus sign, so they're all wrong .

But it's quite probable that there's a mistake in the answer... because 13.79 rad/s2 is exactly half of our answer. As if the dude making the question had forgotten the 1/2 in

$$\theta (t) = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$$

which is a common mistake when one plugs all the numerical values and solve for the answer...