Question

starting at rest t=0,a wheel undergoes a constant angular acceleration .when t=2.33sec, the angular velocity of a wheel is 4.96rad/s.the acceleration continues until t=23.0s,when it abruptly ceases. through what angle does the wheel rotate in the interval t=0 to t=46.0s?

Answer 1

Problem:

Starting at rest $t=0$, a wheel undergoes a constant angular acceleration. When $t=2.33\,\text{sec}$, the angular velocity of a wheel is 4.96 rad/s. The acceleration continues until $t=23.0\,\text{s}$, when it abruptly ceases. Through what angle does the wheel rotate in the interval $t=0$ to $t=46.0\,\text{s}$?

Solution:

According to kinematics of angular movement:

$$\beta = \frac{\omega_1}{t_1} = \frac{4.96}{2.33} = 2.13 \left[ \frac{\text{rad}}{\text{s}^2} \right]$$ $$\omega_2 = \beta t_2 = \omega_1 \frac{t_2}{t_1} = 4.96 \cdot \frac{23}{2.33} = 49 \left[ \text{rad/s} \right]$$ $$\varphi = \varphi_{t_2} + \omega_2 (t_3 - t_2) = \frac{\beta t_2^2}{2} + \omega_2 (t_3 - t_2) = \frac{\omega_1}{t_1} \cdot \frac{t_2^2}{2} + \omega_2 (t_3 - t_2) = 2.13 \cdot \frac{23^2}{2} + 49 \cdot (46 - 23) = 1690 \left[ \text{rad} \right]$$

Where $\beta$ – angular acceleration;

$\omega_1 = 4.96\,\text{rad/s}$ – angular speed at time $t_1 = 2.33\,\text{s}$;

$\omega_2$ – angular speed at time $t_2 = 23\,\text{s}$;

$t_3 = 46\,\text{s}$ – stop time;

$\varphi$ – whole angle, that wheel rotate through time $t_3$;

$\varphi_{t_2}$ – angle, that wheel rotate through time $t_2$;

Answer: $\varphi = 1690$ [rad].