# What is the dimensional formula of centripetal force

Note also that the radial vector r t does not represent the radius of curvature of the path. Time tracking and productivity improvement software with screenshots and website and applications. What affect does a doubling in speed have upon the centripetal force?

Derive the formula for centripetal force F acting on a particle moving in a uniform circle. As we know, the centripetal force acting on a particle moving in a uniform circle depends on its mass m, velocity v and the radius r of the circle.

Hence, we can write. Ans — It is an expression that relates derived quantity to fundamental quantities. But it is not related to the magnitude of the derived quantity. Mass is fundamental quantity but acceleration is a derived quantity and can be represented in terms of fundamental quantities.

The acceleration of any object is defined at the change in its velocity divided by the change in time. Both acceleration and velocity are vectors, so mathematically we can write this as We are going to consider an object moving at a constant speed in a circle, as illustrated in the diagram below. In a time the object has moved through an angleand its initial velocity has changed towhere the only change in the velocity is its direction.

Remember, the velocity is tangential to the radius, so makes a right angle with the radius vector. The direction of the radius vector is, by definition, radially outwards. We will consider an object moving in a circle with a constant speed. It moves through an angle in time. In my blog about vectors I mentioned that, to combine vectors which have different directions, we need to split the vectors into components, add the components and then recombine the resultant components.

The components need to be at right-angles to each other, and usually but not always we choose the x and y-directions when the vectors are in two dimensions. To find the acceleration of our object in this example, we want to find the change or difference in the velocity, that is.

And if the speed of the object is halved decreased by a factor of 2the net force required is decreased by a factor of 4. The mathematical equations presented above for the motion of objects in circles can be used to solve circular motion problems in which an unknown quantity must be determined.

### What is the Dimensional Formula of Force?

The process of solving a circular motion problem is much like any other problem in physics class. The process involves a careful reading of the problem, the identification of the known and required information in variable form, the selection of the relevant equation ssubstitution of known values into the equation, and finally algebraic manipulation of the equation to determine the answer. Consider the application of this process to the following two circular motion problems. Determine the acceleration and the net force acting upon the car.

The solution of this problem begins with the identification of the known and requested information. The solution is as follows:.

The solution is as follows.

Centripetal Force - Frequency and PeriodA kg halfback makes a turn on the formula field. The halfback sweeps out a path that is a portion of the circle with a radius of meters. The halfback makes a quarter of a turn centripetal the circle in 2. Determine the speed, acceleration and net force acting upon the halfback. In Lesson 2 of this unit, circular motion principles and the above mathematical equations will be combined to explain and analyze a variety of real-world motion scenarios including amusement park rides and circular-type motions in athletics.

Anna Litical is practicing a centripetal force demonstration at home. She fills a bucket with water, ties it to a strong rope, and spins it in a circle. Anna spins the bucket when it is half-full of water and when it is quarter-full of water. Expressed using the orbital force T for one revolution of the circle. In particle accelerators, velocity can be very high close to the speed of light in vacuum so the same rest mass now exerts greater inertia relativistic mass thereby requiring greater force for the same centripetal acceleration, so the equation becomes:.

In the case of an object that is swinging around on the end of a rope in a dimensional plane, the centripetal force on the object is supplied by the tension of the rope. The rope example is an example involving a 'pull' force.

## Centripetal force

The centripetal force can also be supplied as a 'push' force, such as in the case where the normal reaction of a wall supplies the centripetal force for a wall of death rider. Newton 's idea of a centripetal force corresponds to what is nowadays referred to as a central force.

### Mathematics of Circular Motion

When a satellite is in orbit around a planetgravity is considered to be a centripetal force even though in the case of eccentric orbits, the gravitational force is directed towards the focus, and not towards the instantaneous center of curvature. Another example of centripetal force arises in the helix that is traced out when a charged particle moves in a uniform magnetic field in the absence of other external forces. In this case, the magnetic force is the centripetal force that acts towards the helix axis. Below are three examples of increasing complexity, with derivations of the formulas governing velocity and acceleration.

Uniform circular motion refers to the case of constant rate of rotation. Here are two approaches to describing this case. Assume uniform circular motionwhich requires three things. While objects naturally follow a straight path due to inertiathis centripetal acceleration describes the circular motion path caused by a centripetal force. The image at right shows the vector relationships for uniform circular motion. In words, the acceleration is pointing directly opposite to the radial displacement r at all times, and has a magnitude:. This result agrees with the previous section, though the notation is slightly different.

When the rate of rotation is made constant in the analysis of nonuniform circular motionthat analysis agrees with this one.

The upper panel in the image at right shows a ball in circular motion on a banked curve. The objective is to find what angle the bank must have so the ball does not slide off the road. Apart from any acceleration that might occur in the direction of the path, the lower panel of the image above indicates the forces on the ball.

There are two forces; one is the force of gravity vertically downward through the center of mass of the ball m gwhere m is the mass of the ball and g is the gravitational acceleration ; the second is the upward normal force exerted by the road at a right angle to the road surface m a n. The centripetal force demanded by the curved motion is also shown above. This centripetal force is not a third force applied to the ball, but rather must be provided by the net force on the ball resulting from vector addition of the normal force and the force of gravity.

The resultant or net force on the ball found by vector addition of the normal force exerted by the road and vertical force due to gravity must equal the centripetal force dictated by the need to travel a circular path.

#### What is Dimensional Formula of Centripetal acceleration ?

The curved motion is maintained so long as this net force provides the centripetal force requisite to the motion. The vertical component of the force from the road must counteract the gravitational force: Substituting into the above formula for F h yields a horizontal force to be:. On the other hand, at velocity v on a circular path of radius rkinematics says that the force needed to turn the ball continuously into the turn is the radially inward centripetal force F c of magnitude:. Consequently, the ball is in a stable path when the angle of the road is set to satisfy the condition:.

If friction cannot do this that is, the coefficient of friction is exceededthe ball slides to a different radius where the balance can be realized. These ideas apply to air flight as well. See the FAA pilot's manual. As a generalization of the uniform circular motion case, suppose the angular rate of rotation is not constant. The acceleration now has a tangential component, as shown the image at right.