Hey, guys. I've got a really fun problem here for you. So the idea here is that we have this ball that's on a cord and we're spinning the ball around in a horizontal circle only. But we're doing this so that the cord always makes a 30-degree angle with the vertical. We want to calculate the ball's speed, its tangential speed as it's going around in a circle. This is our \( V_T \). That's what we're trying to find in this problem.

So how do we do this? Well, we're going to go ahead and stick to the steps. We want to draw a free body diagram. So let's go ahead and do that. We've got the ball like this. We're going to have the weight force that's acting straight down. Let me draw that a little bit bigger. We have the weight force like this. And then we also have no applied forces, no normal or friction, but there is a tension because of the cord. So we have this tension force here like this. Now this tension force is 2-dimensional. So we're going to have to break this up into its x and y components. So we have a \(T_y\) and then we have a \(T_x\). At this point, we know that this \(T_x\) component or the x-component of the cord as the ball is going around in a circle, right, is always going to point towards the inside of the circle. So our \(T_x\) here points like this, and this is responsible for the ball's centripetal acceleration \(\mathbf{a_c = \frac{v^2}{r}}\).

So if we want to find out the \(v_{tangential}\), we're going to have to use \(a_c\) because we know that's related to \(\mathbf{\frac{v^2}{r}}\). Let's consider centripetal direction forces. Remember, our \(T_x\) is the only force that's causing the ball to accelerate centripetally. So \(\mathbf{T_x = m \cdot \frac{v^2}{r}}\). That's the \(v^2\) over \(r\) that we're trying to find.

Let's write out an expression for \(T_x\). This \(T_x\) is going to be related to sine and cosine. So how do we do that? Well, the angle that we were given here is 30 degrees. If we take a look, we kind of just draw out a little right triangle like this. We always want our angles with respect to the horizontal. So the angle we want here is actually 60 degrees. So \(T_x\) is really going to be \(T \cdot \cos(60^\circ)\) and \(T_y\) is going to be \(T \cdot \sin(60^\circ)\).

When we write this \(T_x\), we've got \(\mathbf{T \cdot \cos(60^\circ) = m \cdot \frac{v^2}{r}}\). We're trying to find the velocity but still have other unknowns like the tension and the radius of the circle. The radius would be this distance here, which is \(r\). We will first solve the tension force because we can always look at all the y forces \( \text{sum of all forces in the y-axis} = m \cdot a_y\).

Remember, this ball is only going around in a horizontal plane, meaning the only acceleration that this ball has is in the centripetal direction; it doesn't accelerate vertically. So these forces have to cancel out, meaning \( a_y = 0 \) and \( T_y - mg = 0 \), thus \( T \cdot \sin(60^\circ) = mg \). We can rewrite this equation and solve for \(T\). The expression solves to \( T = \frac{0.5 \cdot 9.8}{\sin(60^\circ)} \), giving a tension of 5.7 N.

Now, to solve for the radius, if we know that the length of the cord is 4 meters and the angle is 30 degrees, we can find another side using the sine rule, which gives the radius as \( 4 \cdot \sin(30^\circ) = 2 \) meters.

Finally, substitute back in, we have \(\mathbf{5.7 \cdot \cos(60^\circ) \cdot 2 \div 0.5}\) for \(v^2\), giving \(v = \sqrt{11.4} = 3.38 \) meters per second. That's how fast this ball is traveling around in its horizontal path as you're spinning it in a circle. Let's move on.