Question

derive an expression for centripetal acceleration and show geometrical e that this acceleration is directed towards the centre of the circle​

Answer 1

Answer:

Centripetal Acceleration is Ac = v2/r

It is directed towards centre

Explanation:

The figure below shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation—the center of the circular path. This direction is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration aca_caca, start subscript, c, end subscript; centripetal means “toward the center” or “center seeking”.

The direction of centripetal acceleration is toward the center of the circle, but what is its magnitude? Note that the triangle formed by the velocity vectors and the triangle formed by the radii r and Δs are similar. Both the triangles ABC and PQR are isosceles triangles with two equal sides. The two equal sides of the velocity vector triangle are the speeds v1=v2=v Using the properties of two similar triangles, we obtain Δv/v=Δs/r

Acceleration is Δv/Δt

Δv=v/rΔs

If we divide both sides by Δt, we get the following:

Δv/Δt=vr×Δs/Δt

Finally, noting that Δv/Δt=ac

we see that the magnitude of the centripetal acceleration is

Ac=v2/r

Direction of centripetal acceleration explained in attached figure

Answer 2

Answer:

The centripetal acceleration is V^2/R

The positive divisors of 8 are 1, 2, 4, and 8. The reciprocals of these numbers, respectively, are 1, 1/2, 1/4, and 1/8. To get the sum, simply add them all: 1 + 1/2 + 1/4 + 1/8. The sum of all the reciprocals of 8 is 15/8 or in mixed form, 1 and 7/8.