Question

A Bestimmen Sie den Beschleunigungsvektor $\overrightarrow{a}(t)$ bei der gleichmäßigen Rotation. Die Parameterdarstellung der Geschwindigkeit ist: $v_x(t)=-\omega r\sin \omega t$ $v_y(t)=\omega r\cos \omega t$ B Der Ortsvektor eines Massenpunktes ist gegeben durch $\overrightarrow{r}(t)=(R\cos \omega t,R\sin \omega t,t).$ Bestimmen Sie die Geschwindigkeit des Massenpunktes zur Zeit $t=\frac{2\pi }{\omega }$. $\mathrm{C}$ Der Beschleunigungsvektor ist beim freien Fall gleich $\overrightarrow{g}=(0,0,-g)$. Wie sieht der Geschwindigkeitsvektor aus, wenn die Geschwindigkeit zur Zeit $t=0$ gleich ${\overrightarrow{v}}_0=(v_0,0,0)$ ist?

Short answer

The acceleration vector for part A) is $(-{\omega }^{2}r\cos (\omega t),{\omega }^{2}r\sin (\omega t))$, the speed vector at time $t=\frac{2\pi }{\omega }$ for part B) is $-R\omega ,0,1$, and the speed vector at time $t$ for part C) is $(v_0,0,-gt)$.

Step-by-step solution

Calculate acceleration vector for part A)

The acceleration vector is calculated as the derivative of the velocity vector $\overrightarrow{v}(t)$. The components of the velocity vector are given as $v_x(t)=-\omega r\sin (\omega t)$ and $v_y(t)=\omega r\cos (\omega t)$. The acceleration vector $\overrightarrow{a}(t)$ thus becomes: ${\overrightarrow{a}}_x(t)=-\frac{d}{dt}(\omega r\sin (\omega t))=-{\omega }^{2}r\cos (\omega t)$ and ${\overrightarrow{a}}_y(t)=\frac{d}{dt}(\omega r\cos (\omega t))={\omega }^{2}r\sin (\omega t)$. This leads to $\overrightarrow{a}(t)=(-{\omega }^{2}r\cos (\omega t),{\omega }^{2}r\sin (\omega t))$.

Calculate velocity for part B)

The location vector $\overrightarrow{r}(t)$ is given as $(R\cos (\omega t),R\sin (\omega t),t)$. To find the speed at the time $t=\frac{2\pi }{\omega }$, we need to first find the velocity vector $\overrightarrow{v}(t)=\frac{d\overrightarrow{r}(t)}{dt}$. This leads to $\overrightarrow{v}(t)=(-R\omega \sin (\omega t),R\omega \cos (\omega t),1)$. At $t=\frac{2\pi }{\omega }$, the velocity becomes $\overrightarrow{v}(\frac{2\pi }{\omega })=(-R\omega ,0,1)$.

Find velocity vector for part C)

In this case, the acceleration vector is equal to the gravitational vector $\overrightarrow{g}=(0,0,-g)$ and the initial speed is ${\overrightarrow{v}}_0=(v_0,0,0)$. Using the kinematic equation $\overrightarrow{v}(t)={\overrightarrow{v}}_0+\overrightarrow{g}t$, we find that the speed at time $t$ is $\overrightarrow{v}(t)=(v_0,0,-gt)$.