5.2.2 Centripetal force

It is important to realise that when an object moves in a circular path it's direction is actually constantly changing meaning it's velocity is constantly changing. A change in velocity means that the object must be accelerating (the greater the change in velocity the greater the acceleration).

We already know that any accelerating object must be under the influence of a net/resultant force.Any force that keeps a body moving with a uniform speed (remember, this is not uniform velocity!) along a circular path is called a centripetal force. A centripetal force is always perpendicular to the velocity of the object meaning that the force has no component in the direction of motion so no work is done on the object. As a result its speed remains constant.

Examples of centripetal force include gravitational attraction for a satellite in orbit, friction for a car going around a corner, and tension in the string of a yo-yo.

At any point on a circular path linear velocity is always constant to the tangent of the path. For an object moving in a circle at a constant speed we can calculate its speed using:

speed = distance/time

We know that in one complete rotation the distance travelled is the circumference of a circle (2πr) and the time is the period (T). This means that:

v = 2πr/T

We know that ω = 2π/T. It therefore follows that:

v = r ω

This means that for objects with the same angular velocity the linear velocity at any instant is proportional to the radius. Increasing the radius will increase linear velocity proportionally.

The acceleration of any object travelling in a circular path at a constant speed is called the centripetal acceleration. It also always acts towards the centre of the circle (like centripetal force). Centripetal acceleration depends on  the speed (v = r ω) of an object and the radius (r) of the circular path. The faster the object travels the larger the acceleration. The smaller the radius is the larger the acceleration. We can determine centripetal acceleration with the following equation:

a = v² / r

NOTE: we can combine this with v = r ω to give a = v = ω² r.

We can combine F = ma with a = v² / r to give an equation for centripetal force:

F = ma = (mv²) / r

This means that for a constant mass and radius the centripetal force (F) is directly proportional to v². Since v = r ω, we can also write the above equation as F = m ω² r. Remember this force is ALWAYS towards the centre of the circular path.

We need to know how to investigate circular motion. What we did in class was effectively to put a bung on the end of a piece of string and spin it around above our heads. Hold the string through a straw and suspend a mass from the other end. The weight will remain stationary is the force it provides (mg) is equal to the centripetal force making the bund travel in a circular path. If the centripetal force required is greater than the weight the mass will move upwards. With this set up we can investigate the centripetal force required for different masses/radii/speeds.

EXTRA:
This stuff isn't specifically in the specification but we covered it in class and it comes up pretty frequently in exam Qs so I thought i'd cover it on the blog. Just skip it if you cba to read lol. It's about sources of centripetal force.

There are many sources of centripetal force, such as tension in a string, gravitational attraction, and friction. We often get questions in exams where the surface is 'banked' (ie sloped) - such as a bike in a velodrome or a plane in the sky. Centripetal force can also be due to changes in the normal contact force when an object travels in a circular path. E.g when a ferris where rotates a net force is required to make you travel in a circular path (as apposed to no net force when your stationary as R = mg).