Question

A particle is moving with constant speed v on a circular path of r radius when it has moved by angle 60°, Find (i) Displacement of particle (ii) Average velocity (iii) Average acceleration?

Answer 1

A) Since the particle has moved by an angle of `60^o`, if we join the initial and final positions of the particles with the center, the figure so formed will be an equilateral triangle. Since all the sides of an equilateral triangle are equal, the displacement would be equal to the radius of the circle.

B)Average velocity of the particle is = total displacent`/`time

Let the radius of the circle be `R`

Displacement = `R`

Now,
total time taken by the particle to complete 1 rotation `i.e` to cover `360^o => T = (2piR)/v`

So, time taken to cover `60^o => T_1 = 1/6T`
`T_1 = (piR)/(3v)`

So, average velocity `= R/((piR)/v)`
`=v/pi`

C)average acceleration = change in velocity`/`time

Let us assume that the particle started with its velocity pointing upwards. ( see the image )

after it has moved by `60^o`, the angle made by its velocity at that instant is `60^o` with the vertical. ( Its just geometry going on here.)

Since we assumed that the initial velocity was in upwards direction, the angle made by the final velocity with the initial velocity is indeed `60^o`.

Since we are finding the change in the velocity,
Let final velocity be `v_2` and initial `v_1`

`= DeltaV = sqrt((v_1)^2 + (v_2)^2 - 2v_1v_2(cos(60^o)`

Since the particle hasn't changed its velocity, `v_1 = v_2`

Solving the above expression we get

`= DeltaV = v_1`

Plugging in the values,

average velocity `= (3v^2)/(piR)`