This is the force that keeps a body moving with uniform speed in a circular path. It is the inward internal force that is required for a body to maintain uniform speed in a circular or curved path. The acceleration of this force is usually directed towards the center of the circle. The acceleration of a body undergoing circular motion is called centripetal acceleration. A practical example of Centripetal force is turning a car along a roundabout.
The magnitude of the centripetal acceleration is given by:
$$ a = \frac{v²}{r} $$
$$ \text{Also, } F = ma $$
$$ \text{centripetal force }{F_T} = \frac{mv²}{r} $$
$$ \text{where m = mass in kg} $$
$$ \text{v = velocity in m/s} $$
$$ \text{r = radius in m} $$
$$ {F_T} \text{ is in Newtons} $$
Note: The direction of centripetal acceleration is usually perpendicular to the direction of the velocity.
According to the Newton's third law of motion, action and reaction are equal and opposite. Hence a body undergoing uniform circular motion experiences a force opposite the centripetal force acting on that body. Centrifugal force is the apparent force acting on a body along a circular path whose acceleration moves away from the centre of the circle. Centrifugal force is the force that causes a passenger in a car moving in a circular path to move away to the other side of the car. $$ \text{Centrifugal force,F} = -\frac{mv²}{r} $$
| Centripetal Force | Centrifugal Force |
|---|---|
| Acts towards the center of rotation. | Acts away from the center of rotation. |
| Required for circular motion. | Perceived force in a rotating frame of reference. |
| Causes objects to move in a circular path. | Apparent force due to inertia in a rotating system. |
| Provides the necessary acceleration. | Not a "real" force but a perceptual effect. |
| Examples include tension, gravitational force. | Examples include the feeling of being pushed outward in a turning car. |
| Always directed towards the center of the circle. | Always directed away from the center of the circle. |
Angular velocity is the rate of change of angular displacement with respect to time. It is typically denoted by \( \omega \) (omega) and measured in radians per second (rad/s).
$$ \omega = \frac{\theta}{\text{t}} $$ $$ \text{But, } \theta = \frac{s}{t} $$ $$\text{where }\theta \text{ is in radians} $$ $$ \text{s is in m} $$ $$ \text{r is in m} $$ $$ \theta \text{ is in rad/s} $$ $$ \therefore s = r\theta$$ $$ \text{Also, v } = \frac{s}{t} $$ $$ \therefore v = \frac{r × \theta}{t} $$ $$ v = \omega r $$ $$ {F_T} = \frac{m{(\omega r)}^2}{r} $$ $$ {F_T} = m\omega² r $$ $$ a = \omega² r $$Example 1: A body moving along a circular path with uniform angular speed of 0.6rad/s and at a constant speed of 3m/s. Calculatate the acceleration of the body towards the center of the circle.(WAEC)
Example 2: A stone tied to a string is made to revolve in a horizontal circle of radius 4m with an angular speed of 2 rad/s. With what tangential velocity will the stone move off the circle if the string cuts. (WAEC)
When the stone is cut from the string, it moves with a linear speed,v according to the law of inertia. Hence;
$$ v = \omega r $$ $$ r = 4m $$ $$ \omega = 2 rad/s $$ $$ v = 2 × 4 = 8m/s $$Example 3: An object of mass 5kg moves round a circle of radius 6m. If the period of the oscillation is πs. Calculate the magnitude of the centripetal force on the body. (NECO)
Since the body moved round the circle it has completed one revolution = 360⁰ = 2π
$$ mass = 5kg $$ $$ radius = 6m $$ $$ t = ? $$ $$ \text{Period, T} = πs $$ $$T = \frac{\text{time taken, t}}{\text{number of oscillations, n}} $$ $$ π = \frac{t}{1} $$ $$ t = πs $$ $$ \omega = \frac{\theta}{t} $$ $$ \theta = 360^\circ = 2π $$ $$\omega = \frac{2π}{π} $$ $$ \omega = 2rad/s $$ $$ {F_T} = m\omega²r $$ $$ = 5 × 2² × 6 = 120N $$