Centripetal acceleration appears at the movement of a body on a circle. Directionally it to its center, is measured in m/s². Feature of this type of acceleration is that it is even then when the body moves with a constant speed. It depends on the radius of a circle and linear speed of a body.

## It is required to you

- - speedometer;
- - device for measurement of distance;
- - stop watch.

## Instruction

1. To find centripetal acceleration, measure the speed of the body moving on circular trajectories. It is possible to make it by means of a speedometer. If it is not possible to install this device, calculate linear speed. For this purpose note time which was spent for a whole revolution on a circular trajectory.

2. This time is the rotation period. Express it in seconds. Measure the radius of a circle on which the body a ruler, a roulette or a laser range finder in meters moves. To find speed find the work of number 2 on number π ≈ 3.14 and radius R circles and divide result for T. It will also be the linear speed of a body of v=2 ∙π ∙ R/T.

3. Find centripetal acceleration ats, having divided a square of linear speed of v into the radius of a circle on which R body moves (ats =v²/R). Using formulas for determination of angular speed, frequency and the period of rotation, you find this size and on other formulas.

4. If angular speed ω is known, and the trajectory radius (on which circle the body moves) R that centripetal acceleration will be equal ats = ω² ∙ to R. When T, and R trajectory radius, ats = 4 ∙π² ∙ is known the period of rotation of a body to R/T². If rotation frequency ν is known (the number of full rotations in one second), then determine centripetal acceleration by a formula ats = 4 ∙π² ∙ by R ∙ν².

5. Example: the car which radius of wheels 20 cm, move on the road with a speed of 72 km/h. Define centripetal acceleration of extreme points of its wheels. Decision: linear speed of points of any wheel will be 72 km/h =20 m/s. Transfer the radius of a wheel to m R=0.2 meters. Calculate centripetal acceleration, having substituted the obtained data in a formula ats to =v²/R. Receive ats =20²/0.2=2000 m/s². At the uniform rectilinear motion, extreme points of all four wheels of the car will have this centripetal acceleration.