Velocity and Acceleration — Question 4

Question 4 – Solution

Strategy. Differentiate symbolically and use trigonometric identities and dot products.

Step 1: Differentiate $$\mathbf v=\left\langle -R\omega\sin(\omega t),R\omega\cos(\omega t),0\right\rangle,$$ so $$\boxed{\|\mathbf v\|=R\omega}.$$ Differentiating again, $$\boxed{\mathbf a=\left\langle -R\omega^2\cos(\omega t),-R\omega^2\sin(\omega t),0\right\rangle}.$$

Step 2: Orthogonality $$\begin{align*} \mathbf v\cdot\mathbf a &=R^2\omega^3\sin(\omega t)\cos(\omega t) -R^2\omega^3\cos(\omega t)\sin(\omega t)=0. \end{align*}$$ Thus $\boxed{\mathbf v\perp\mathbf a}$.

Step 3: Inward acceleration Comparing with $\mathbf r$ gives $$\boxed{\mathbf a(t)=-\omega^2\mathbf r(t)}.$$ It points toward the origin, and $$\boxed{\|\mathbf a\|=R\omega^2=\frac{(R\omega)^2}{R}=\frac{v^2}{R}}.$$ This is purely normal acceleration; the constant speed confirms there is no tangential component.