Question

object is moving on a circular track of radius R with constant speed V the average acceleration of the object during time interval in which it complete one fourth revolution of the track will be​

Answer 1

Answer:

$\frac{2\sqrt{2}v^2}{\pi R}$

Explanation:

1) Consider coordinate system as shown in figure.

Initial velocity and final velocity ,

$u_i = vj\\ \\ u_f=-vi$

2) Then, for time taken :

For One-fourth of revolution

Total distance covered : $D=\frac{\pi R}{2}$

Speed , = v

Hence, Time taken will be

$T = \frac{D}{v} =\frac{\pi R}{2v}$

3) Average Acceleration :

$a_{avg}=\frac{\Delta v }{\Delta t } =\frac{|-vi-vj|}{\frac{\pi R}{2v}}\\ \\ =\frac{\sqrt{v^2+v^2}*2v}{\pi R} \\ \\=\frac{2\sqrt{2}v^2}{\pi R}$

Answer 2

The average acceleration of the object during time interval in which it complete one fourth revolution of the track will be $\frac{2\sqrt{2}V^2}{\pi R}$

Explanation:

Radius of track=R

Speed of object=V

Distance covered by object in one revolution=Circumference of circle$2\pi R$

Distance covered by object=$\frac{1}{4}\times 2\pi R=\frac{\pi R}{2}$

Initial velocity and final velocity

$u_i=v_j=V$

$u_f=-v_i$

$v_i=V$

Time taken by object=$\frac{Distance}{velocity}$

Time taken by object=$\frac{\pi R}{2V}$

Average acceleration=$\frac{change\;in\;velocity}{time}$

Average acceleration=$\frac{\mid{-v_i-v_j}\mid }{\frac{\pi R}{2V}}$

Average acceleration=$\frac{\sqrt{v^2_i+v^2_j}\times 2V}{\pi R}$

Average acceleration=$\frac{\sqrt{V^2+V^2}\times 2V}{\pi R}$

Using formula : Magnitude=$\sqrt{x^2+y^2}$

Where x=Vector component along x-axis

y=Vector component along y- axis

Average acceleration=$\frac{2\sqrt{2}V^2}{\pi R}$

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https://brainly.in/question/4776850 Answered by Siril