Centripetal Forces - Online Tutor, Practice Problems & Exam Prep

Centripetal forces are essential in understanding uniform circular motion, where acceleration always points toward the center of the circle. The net force causing this acceleration can be analyzed using the equation $F=ma$, with centripetal acceleration defined as $a=v^2/r$. In practical problems, drawing a free body diagram helps identify forces, such as tension, which is crucial for maintaining circular motion.

Intro to Centripetal Forces

Guys, now that we've covered the basics of uniform circular motion, it's time to start talking about forces. When you have forces in circular motion, those are called centripetal forces. The whole idea with this video is that the way we have solved forces problems in the past, we had forces along straight lines, these are called linear force problems, is going to be the same way that we solve circular or centripetal forces problems. But there are a couple of differences I want to go through. So, let's do a quick example and check this out.

To recap, whenever we had linear forces problems, we had forces along fixed x and y axes. We drew our coordinate system like this, and let's say we had a couple of forces in the x axis for simplicity. If we had multiple forces, then basically the net force is going to produce an acceleration. You have an acceleration and force those point along the same directions. And the way that we solve this is just by using F=ma in the x and y axes. Now let's talk about circular motion. We know that in circular motion, you have some tangential velocity and your acceleration always points towards the center of the circle. The same principle applies. If you're accelerating towards the center, there has to be some net force that's pushing you towards the center. The difference is that depending on where you are in the path, those directions are going to change. It actually doesn't really make sense for us to use a fixed x and y coordinate system because the directions are always changing in circular motion. So what we do is we just have forces along the centripetal direction, which really just means that wherever you are on the path, those forces are going to be pointing toward or away from the center. But ultimately, we still have the same variables, F_net and A_c. So, we're just going to use F=ma to solve these kinds of problems. The only difference is that we're going to use the sum of all forces in the centripetal axis equals mass times acceleration centripetal. That's really all there is to it. It's still F=ma, but now everything is just in circular motion. The way that we solve these problems is actually very similar to how we solve problems before. We're going to draw our free body diagram and then write F=ma.

Now, let's consider a problem where we have this 3-kilogram block that's tied to a string, and it slides around the circle. Essentially, this block is going to be sliding around in a circle on this horizontal tabletop. We want to calculate the tension on the string. To avoid confusion with multiple variables involving 'T' (as it also represents the period), we're going to solve for the force of tension. We're going to start off with the weight force. Suppose we draw a side view as if looking at the tabletop from the side. You would see the tabletop, the box, and the weight force that's straight down (your mg), and then you would have the force of tension caused by the string, and then you'd also have a normal force.

If you were looking at the tabletop from the top, you would just see the path of the circle. At any point, the tension force points towards the center. Our forces to consider are the normal, the weights, and the tension. But the normal and the weight force are only acting along the vertical plane. When you look at it from the top, the only thing that's actually keeping this object moving in a circle is really just the force of tension. So here's our ft, and that's the force of tension. When we expand our F=ma, the only force that's actually causing this to accelerate centripetally is our force of tension. Then we express this acceleration centripetal, A_c, as v2/r which is crucial.

In this example, the block completes a rotation every 4 seconds, allowing us to calculate that v is 2πrt. If we calculate using the given radius and period, we get a velocity of 3.14 m/s. Plugging this back into mv2/r, we get a tension force of 14.8 newtons. Just like that, using the same steps of free body diagram and F=ma, but now with additional equations involving the centripetal acceleration that we might have to solve for.

A small 4kg block is tied to the end of 3m string and slides around in a circle on a frictionless table. Suppose the string will break if the tension exceeds 50N. Find the maximum speed the block can have without breaking the string.

• A. 37.5 m/s
• B. 6.1 m/s
• C. 8.7 m/s
• D. 2.8 m/s

Conical Pendulum

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.

Here’s what students ask on this topic:

What is centripetal force and how is it calculated?

Centripetal force is the net force that causes an object to move in a circular path, always pointing towards the center of the circle. It is essential for maintaining uniform circular motion. The centripetal force $F$_c can be calculated using the equation:

$F$_c = ma_c, where $m$ is the mass of the object and $a$_c is the centripetal acceleration. The centripetal acceleration $a$_c is given by:

$a$_c = v2r, where $v$ is the tangential velocity and $r$ is the radius of the circular path.

How do you draw a free body diagram for an object in circular motion?

To draw a free body diagram for an object in circular motion, follow these steps:

1. Identify the object and the forces acting on it.
2. Draw the object as a point or a simple shape.
3. Represent each force with an arrow pointing in the direction the force is acting. Label each force (e.g., tension $T$, gravitational force $\mathrm{mg}$, normal force $N$).
4. For circular motion, ensure that the centripetal force points towards the center of the circle.

For example, if a block is tied to a string and moves in a horizontal circle, the tension in the string provides the centripetal force. The free body diagram would show the tension force pointing towards the center of the circle, while the gravitational and normal forces act vertically.

What is the relationship between centripetal force and centripetal acceleration?

The relationship between centripetal force and centripetal acceleration is given by Newton's second law of motion. The centripetal force $F$_c is the net force causing centripetal acceleration $a$_c in an object moving in a circular path. This relationship is expressed as:

$F$_c = ma_c, where $m$ is the mass of the object. The centripetal acceleration $a$_c is given by:

$a$_c = v2r, where $v$ is the tangential velocity and $r$ is the radius of the circular path. Thus, the centripetal force is directly proportional to the mass of the object and the square of its velocity, and inversely proportional to the radius of the circular path.

How do you calculate the tension in a string for an object in circular motion?

To calculate the tension in a string for an object in circular motion, follow these steps:

1. Draw a free body diagram to identify the forces acting on the object.
2. Use the centripetal force equation: $F$_c = ma_c.
3. Express centripetal acceleration as $a$_c = v2r.
4. Substitute the known values (mass $m$, velocity $v$, and radius $r$) into the equation.

For example, if a 3 kg block is tied to a string and moves in a circle with a radius of 2 m and a velocity of 3.14 m/s, the tension $T$ in the string is calculated as:

$T=m\frac{v^2}{r}$

Substituting the values:

$T=\; 3\frac{2^{}}{}=\; 14.8\; N$

What is the difference between centripetal and centrifugal force?

Centripetal force is the real force that acts on an object moving in a circular path, directing it towards the center of the circle. It is responsible for maintaining the object's circular motion. The equation for centripetal force is:

$F$_c = mv2r

On the other hand, centrifugal force is a fictitious force that appears to act on an object moving in a circular path when observed from a rotating reference frame. It seems to push the object away from the center of the circle but does not exist in an inertial frame of reference. Centrifugal force is often perceived due to the inertia of the object resisting the centripetal force.