Question

A tub of a washer start from rest and reach an angular speed of 5 rev/s in 8s. Then, tub slows to rest in 12s. through how many revolutions does the tub turn during the entire 20 s interval? Angular acceleration is constant during starting and stopping.

Answer 1

A tub of a washer start from rest and reach an angular speed of 5 rev/s in 8s. Then, tub slows to rest in 12s. through how many revolutions does the tub turn during the entire 20 s interval? Angular acceleration is constant during starting and stopping

1) For starting:

Angular acceleration equals:

$$\beta_{start} = \frac{\Delta \omega_{start}}{\Delta t_{start}}$$

$\Delta \omega_{start}$ – changing of angular velocity

$$\Delta t_{start} - \text{time}$$ $$\beta_{start} = \frac{5 \text{ rev}}{s} = \frac{s}{8} \text{ rev}/s^2$$

Number of revolutions for motion with constant angular acceleration equals:

$$N_{start} = \frac{\beta_{start} \Delta t_{start}^2}{2} = \frac{5}{8} \cdot 8^2 = 5 \cdot \frac{8}{2} = 20$$

2) For stopping:

Angular acceleration equals:

$$\beta_{\text{stop}} = \frac{\Delta \omega_{\text{stop}}}{\Delta t_{\text{stop}}}$$

$\Delta \omega_{\text{stop}}$ – changing of angular velocity

$$\Delta t_{\text{stop}} - \text{time}$$ $$\beta_{\text{stop}} = \frac{5 \text{ rev}}{s} = \frac{5}{12} \text{ rev}/s^2$$

Number of revolutions for motion with constant angular acceleration equals:

$$N_{\text{stop}} = \frac{\beta_{\text{stop}} \Delta t_{\text{stop}}^2}{2} = \frac{5}{12} \cdot 12^2 = 5 \cdot \frac{12}{2} = 30$$

Finally, total number of revolutions equals:

$$N_{\text{total}} = N_{\text{start}} + N_{\text{stop}} = 20 + 30 = 50$$

Answer: 50.