Answer to Question #211804 in Mechanics | Relativity for Nash

Question

A wheel rotating about a fixed axis with a constant angular acceleration of 1.0rad/s2 turns through 1.5 revolutions during a 5.0s time interval. Calculate the angular velocity at the end of this time interval. 

Answer 1

We can calculate the angular velocity at the end with the use of the angular acceleration

($\alpha=1\frac{rad}{s^2}=cte$) and the initial angular velocity $\omega_0$ with the frequency of spin given

(1.5 revolutions during a 5.0 s time interval):

$\omega_0 = 2\pi f=(2\pi \frac{rad}{\cancel{rev}})(\frac{1.5\,\cancel{rev}}{5\,s})=0.6\pi \frac{rad}{s}\approxeq 1.8849\,\frac{rad}{s}$

$\omega_f=\omega_0+\alpha t=1.8849\,\frac{rad}{s}+(1\dfrac{rad}{s^{\cancel{2}}})(5\,\cancel{s})$

$\implies \omega_f=6.8849\,\frac{rad}{s}$

In conclusion, the angular velocity at the end of the time interval is 6.8849 rad/s.

Reference:

• Serway, R. A., & Jewett, J. W. (2018). Physics for scientists and engineers. Cengage learning.