# Derivation of Rotational Motion Equations using Calculus

How are the equations for rotational motion derived using calculus and the following general equations ?

$$\mathbf{v}(t) = \mathbf{v}_0+\int_{t_0}^t \mathbf{a}(t')dt'$$

$$\mathbf{r}(t) = \mathbf{r}_0+\int_{0}^t \mathbf{v}(t')dt'$$

Let $a(t) = \alpha$

Also, can the polar unit vectors be present throughout the answer so I can see how they effect integration $(\mathbf{r},\theta)$.

• I edited your question to use Mathjax which is the site standard for mathematical expressions. We actively discourage images of equations. – StephenG Aug 3 '18 at 3:00

Calculate the positions $r(t)$ of any three distinct, non-coplanar points A, B, C in the rigid body. As the object translates and/or rotates, all other points in the rigid body maintain the same positions relative to A, B, C.

If $\alpha$ is a constant then the orientation of the object in 2D is $\theta=\frac12 \alpha t^2$.

You don't need those general linear-motion definitions/equations. Rather, you need the equivalent rotational-motion definitions/equations:

$$\mathbf{\omega}(t) = \mathbf{\omega}_0+\int_{t_0}^t \mathbf{\alpha}(t')dt'\\ \mathbf{\theta}(t) = \mathbf{\theta}_0+\int_{0}^t \mathbf{\omega}(t')dt'$$

These are fundamental definitions that just mathematically reflect the fact that angular velocity is the change in angular position, and angular acceleration the change in angular velocity, in the same way as in the linear case.

The four usual motion equations are derived by assuming $$\alpha$$ constant, and in exactly the same way as the linear motion equations are derived. See for example here.

Your question is a bit vague, so the other answers might not be satisfactory- or maybe this won't be either.

If you are asking about where the relationship between angular quantities and linear quantities comes from:

For a simple circle (something which all curves can be approximated to), the arc length is give by $$l=\Delta\theta \cdot R$$ differentiating both sides gives (assuming the radius is time-independent): $$v_{tangential}=\dot{\theta } \cdot R$$ If the radius is time-dependent, according to the product rule: $$v_{tangential}=\dot{\theta} \cdot R +\theta \cdot \dot{R}$$

You can substitute in your formulae for velocity and radius to get the angular velocity as a function of time. You can also differentiate this again to obtain an expression for (tangential) acceleration.