Constant Angular Acceleration

[blue5t1053]

Question:
A flywheel has a constant angular acceleration of 2 rad/sec^2. During the 19 sec time period from t1 to t2 the wheel rotates through an angle of 15 radians. What was the magnitude of the angular velocity of the wheel at time t1?

Hint: let t1=0 sec, and t2=t

Equations:
$$\vartheta - \vartheta_{0} = \omega_{0} t + \alpha t ^{2}$$

My Work:
$$(15 radians) - (0 radians) = \omega_{0} (19 sec) + (2 \frac{rad}{sec^{2}})(19 sec )^{2}$$

$$(15 radians) - (2 \frac{rad}{sec^{2}})(19 sec )^{2} = \omega_{0} (19 sec)$$

$$\frac{(15 radians) - (2 \frac{rad}{sec^{2}})(19 sec )^{2}}{(19 sec)} = \omega_{0}$$

$$\omega_{0} = (-18.2)\frac{rad}{sec} = (18.2)\frac{rad}{sec} \ for \ magnitude; \ at \ t1$$

Did I do everything right?

 

[quantumdude]

Question:
A flywheel has a constant angular acceleration of 2 rad/sec^2. During the 19 sec time period from t1 to t2 the wheel rotates through an angle of 15 radians. What was the magnitude of the angular velocity of the wheel at time t1?

Hint: let t1=0 sec, and t2=t

Equations:
$$\vartheta - \vartheta_{0} = \omega_{0} t + \alpha t ^{2}$$

My Work:
$$(15 radians) - (0 radians) = \omega_{0} (19 sec) + (2 \frac{rad}{sec^{2}})(19 sec )^{2}$$

$$(15 radians) - (2 \frac{rad}{sec^{2}})(19 sec )^{2} = \omega_{0} (19 sec)$$

$$\frac{(15 radians) - (2 \frac{rad}{sec^{2}})(19 sec )^{2}}{(19 sec)} = \omega_{0}$$

You're good up to here.

$$\omega_{0} = (-18.2)\frac{rad}{sec} = (18.2)\frac{rad}{sec} \ for \ magnitude; \ at \ t1$$

You've miscalculated. You should get something close to 37 rad/s for the magnitude.

 

[manjuvenamma]

I think your equation is wrong

Should it not be

theta(final) - theta(initial) = time * angular velocity(initial) + (1/2) * angular acceleration * time^2.

similar to the equation in linear motion?

 

[Doc Al]

I think your equation is wrong

You are right.

Equations:
$$\vartheta - \vartheta_{0} = \omega_{0} t + \alpha t ^{2}$$

That should be:
$$\vartheta - \vartheta_{0} = \omega_{0} t + (1/2)\alpha t ^{2}$$