# How can I find the tangential speed of an object from vectors when there is angular acceleration?

The problem is as follows:

A protein sample spins in the counterclockwise direction in a centrifuge seen from the top as shown in the diagram from below. The radius of the centrifuge es $$R=2\,m$$. The magnitude of its speed changes. At a certain instant the acceleration vector is as shown in the figure. Find the speed in $$\frac{m}{s}$$ and state the type of its motion in the given instant. A for acceleration if the speed increases or D for deceleration if the speed decreases. The alternatives given on my book are:

$$\begin{array}{ll} 1.10\frac{m}{s};\,A\\ 2.10\frac{m}{s};\,D\\ 3.5\frac{m}{s};\,A\\ 4.5\frac{m}{s};\,D\\ 5.10\frac{m}{s};\,\textrm{uniform motion}\\ \end{array}$$

This problem I'm particulary lost at. The acceleration shown in the graph. What is it?. Is it perhaps the total acceleration?. In other words the norm of the centripetal and the tangential acceleration?

If so then that would meant that the:

$$a_c=50\frac{m}{s^2}$$

Then the tangential acceleration will be:

$$a_t=50\frac{m}{s^2}$$

And because the angular acceleration is related to the tangential acceleration due the radius as:

$$a_t=\alpha\times r$$

Then:

$$\alpha=\frac{a_t}{r}=\frac{50}{2}=25\frac{rad}{s^2}$$

But that's how far I went in my analysis. What else can I relate to find the asked speed?.

The only equations which I recall are:

$$\omega_{f}=\omega_{0}+\alpha t$$

Can somebody help me here?. What exactly should be the right path to get the answer?.

• You're probably better off asking this on the physics.stackexchange.com forum. Nov 4, 2019 at 1:33
• Centripetal acceleration is $a_c=v^2/R$ that gives you connection between velocity and acceleration. Please, see en.wikipedia.org/wiki/Centripetal_force for more details. Nov 4, 2019 at 1:41

The centripetal acceleration is $$50 \frac{m}{s^2}$$; as @guest stated, this leads to a velocity of $$\sqrt{50 \frac{m}{s^2} \cdot 2m} = 10 \frac{m}{s}$$.
• No. They are completely independent. However, $\vec{a} = \vec{a}_c + \vec{a}_t$. Nov 4, 2019 at 8:27
You have to decompose the acceleration vector into two components, one which is tangential to the orbit and the other which points towards the center of the circle. Note that the tangential component is paralel to the velocity $$\vec{v}$$ but it points in the opposite direction , so there is gonna be deceleration. The component pointing towards the center is the centripetal acceleration given by $$a_{cp} = \frac{v^{2}}{R}$$. Remember each component of the acceleration can be calculated using pythagoras theorem.
• They are completely independent. A centripetal acceleration is needed to keep the particle in a circular trajectory. It has nothing to do with the tangential motion: the particle could be accelerating tangentially or not. However, note that the general formula for the centripetal acceleration is $a_{cp}=v^{2}/R$ so if the particle is supposed to keep its distance from the center of the circle, then an increase of $v$ should imply an increase on $a_{cp}$ and, as a consequence, an increase on the centripetal force $F_{cp}=ma_{cp}$. Nov 4, 2019 at 11:50