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JinM
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[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.
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
\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
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 \thetaI'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
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 \thetaI'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
Last edited:
