Answer to Question #134733 in Physics for esra

Question

A wheel of 15 mm radius on a rotates on a stationary axle . it is uniformly Speed ed up from rest to a speed 500 rpm in a time of 30 s. Find.1- The angular acceleration of the wheel .2- The tangential acceleration of a point on its rim.3- The average angular speed .4- The angular displacement .5- The length of wheel.

Answer 1

1) We can find the angular acceleration of the wheel from the kinematic equation:

$$\omega_f = \omega_i + \alpha t,$$

here, $\omega_i = 0 \ \dfrac{rad}{s}$ is the initial angular velocity of the wheel, $\omega_f = (500 \ \dfrac{rev}{min}) \cdot (2 \pi \dfrac{rad}{1 \ rev}) \cdot (\dfrac{1 \ min}{60 \ s}) = 52.36 \ \dfrac{rad}{s}$ is the final angular velocity of the wheel, $\alpha$ is the angular acceleration of the wheel and $t = 30 \ s$ is the time during which the wheel accelerates.

Then, from this formula we can calculate the angular acceleration of the wheel:

$$\alpha = \dfrac{\omega_f - \omega_i}{t} = \dfrac{52.36 \ \dfrac{rad}{s} - 0 \ \dfrac{rad}{s}}{30 \ s} = 1.74 \ \dfrac{rad}{s^2}.$$

2) We can find the tangential acceleration of the point on its rim from the formula:

$$a_t = r \alpha,$$ $$a_t = 15 \ mm \cdot (\dfrac{1 \ m}{1000 \ mm}) \cdot 1.74 \ \dfrac{rad}{s^2} = 0.0261 \ \dfrac{m}{s^2}.$$

3)-4) Let's first find the angular displacement from the kinematic equation:

$$\theta = \omega_it + \dfrac{1}{2} \alpha t^2,$$ $$\theta = 0 + \dfrac{1}{2} \cdot 1.74 \ \dfrac{rad}{s^2} \cdot (30 \ s)^2 = 783 \ rad.$$

Then, we can find the average angular speed from the formula:

$$\omega = \dfrac{\theta}{t} = \dfrac{783 \ rad}{30 \ s} = 26.1 \ \dfrac{rad}{s}.$$

5) We can find the length of the wheel from the formula:

$$l =2 \pi r = 2 \cdot \pi \cdot 0.015 \ m = 0.094 \ m.$$

Answer:

1) $\alpha = 1.74 \ \dfrac{rad}{s^2}.$

2) $a_t = 0.0261 \ \dfrac{m}{s^2}.$

3) $\omega = 26.1 \ \dfrac{rad}{s}.$

4) $\theta = 783 \ rad.$

5) $l = 0.094 \ m.$