Question

A wheel rotates with a constant angular acceleration of 3.50 rad/s2. (a) If the angular
speed of the wheel is 2.80 rad at t = 0, through what angular displacement does the
wheel rotate in 2.5 sec? (b) Through how many revolutions has the wheel turned during
this time interval? (c) What is the angular speed of the wheel at t = 2.5 sec?

Answer 1

(a) The angular diaplacement is given by the kinematic law:

$$\phi = \omega_0t + \dfrac{\varepsilon t^2}{2}$$

where $\omega_0 = 2.8 \space rad/s$ is the initial anglar speed, $\varepsilon = 3.5 \space rad/s^2$ is the constant angular acceleration, and $t = 2.5s$ is a time.

Hence, obtain:

$$\phi = 2.8\cdot 2.5 + \dfrac{3.5\cdot 2.5^2}{2} = 17.9375 \space rad$$

(b) One revolution takes $2\pi$ rad. Thus, the number of revolutions will be:

$$n = \dfrac{\phi}{2\pi} = \dfrac{17.9375}{2\pi} \approx 2.85$$

(c) By definition, the angular speed is given by the following expression (for constant acceleration):

$$\omega = \omega_0 + \varepsilon t$$

Hence, obtain:

$$\omega = 2.8 + 3.5\cdot 2.5 = 12.6 \space rad/s$$

Answer. (a) 17.9375 rad, (b) 2.85 revolutions, (c) 12.6 rad/s.