Question

Starting from rest at time t = 0, a grindstone has a constant angular acceleration of 3.2rad/s². What is the angular displacement after 2.7 s? If the radius of the grindstone is 0.24 m. Calculate the tangential and radial acceleration.

Answer 1

Starting from rest at time $t = 0$, a grindstone has a constant angular acceleration of $3.2\mathrm{rad} / \mathrm{s}^2$. What is the angular displacement after 2.7 s? If the radius of the grindstone is 0.24 m. Calculate the tangential and radial acceleration.

Angular displacement after 2.7 s:

$$\varphi = \frac {\beta t ^ {2}}{2}$$ $$\varphi = \frac {3.2\mathrm{rad} / \mathrm{s} ^ {2} * (2.7\mathrm{s}) ^ {2}}{2} = 11.66\mathrm{rad}$$

First of all let's find an angular speed:

$$\omega = \beta t$$

Where $\beta$ - angular acceleration.

$$\omega = 3.2\mathrm{rad} / \mathrm{s} ^ {2} * 2.7\mathrm{s} = 8.64\mathrm{rad} / \mathrm{s}$$

Centripetal acceleration:

$$a _ {n} = \omega^ {2} R = (8.64\mathrm{rad} / \mathrm{s}) ^ {2} * 0.24\mathrm{m} = 17.92\mathrm{m} / \mathrm{s} ^ {2}$$

Tangential acceleration:

$$a _ {t} = \beta R$$ $$a _ {t} = 3.2\mathrm{rad} / \mathrm{s} ^ {2} * 0.24\mathrm{m} = 0.77\mathrm{m} / \mathrm{s} ^ {2}$$

Answer: $\varphi = 11.66\mathrm{rad}$, $a_{n} = 17.92\mathrm{m} / \mathrm{s}^{2}$, $a_{t} = 0.77\mathrm{m} / \mathrm{s}^{2}$