Different values with rotational kinematic equations

[JinM]

[SOLVED] Different values with rotational kinematic equations

Hello everyone,
I got two different values for final angular speed when I tried to use the third and fourth kinematic equations. I filled the template here with the problem with my attempts.

Homework Statement

A bicycle wheel of radius r = 1.5 m starts from rest and rolls 100 m without slipping in 30 s. Assuming that the angular acceleration of the wheel given above was constant, calculate: a) The angular acceleration, b) the final angular velocity c) the tangential velocity and tangential acceleration of a point on the rim after one revolution.

Given

$$\Delta\theta = 100/1.5 = 66.7$$ rad

$$\omega_{i} = 0$$

$$\Delta t = 30 s$$

$$\alpha$$ is constant

Unknowns

$$\alpha = ?$$

$$\omega = ?$$

$$v_{t} = ?$$ and $$a_{t} = ?$$ when $$\Delta\theta = 2\pi$$

Homework Equations

$$\Delta\theta = \frac{1}{2} (\omega_{f} + \omega_{i}) \Delta t$$

$$\omega_{f}^{2} = \omega_{i}^2 + 2 \alpha \Delta \theta$$

$$\omega_{f} = \omega_{i} + \alpha \Delta t$$

The Attempt at a Solution

Part b:

$$\Delta\theta = \frac{1}{2} (\omega_{f} + \omega_{i}) \Delta t$$

$$66.7 = \frac{1}{2}(0 + \omega_{f})(30)$$

$$\omega_{f} = 4.45$$ rad/s

Part a:

$$\omega_{f} = \omega_{i} + \alpha \Delta t$$

$$4.45 = 0 + 30\alpha$$

$$\alpha = 0.15$$ rad/s^2

Part c:

My strategy is to find the final angular velocity $$\omega_f$$ using the first relevant equation. Then finding the tangential velocity by multiplying $$\omega_f$$ with the radius.

$$\Delta\theta = \frac{1}{2} (\omega_{f} + \omega_{i}) \Delta t$$

$$2\pi = \frac{1}{2} (\omega_{f} + 0)(30)$$

$$\omega_{f} = \frac{2\pi}{15} = 0.419$$

And then:

$$v_{t} = r\omega = (1.5)(0.419) = 0.6285$$ m/s

$$a_{t} = r\alpha = (1.5)(0.15) = 0.255$$ m/s^2However, the solution key has a different answer. $$\omega_{f}$$ after one revolution equals 1.37 rad/s and $$v_{t}$$ = 2.06. Apparently, they used this equation:

$$\omega_{f}^{2} = \omega_{i}^2 + 2 \alpha \Delta \theta$$I'm a little confused. I double checked my calculations, and I am almost sure that I haven't made a calculation error. I appreciate it if someone could help me with this.

Thanks,
Jin

 

[D H]

\pi[/tex] = 1/2 * (0 + $$\omega$$f) * 30

Your mistake is assuming that the time here is 30 seconds. It takes quite a bit less than 30 seconds for the wheel to make one revolution.

 

[JinM]

OH! I get it now. Thanks a lot, D H