Question

At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (7.00 m/s2)i hat + (2.00 m/s2)j. It moves at constant speed. At time t2 = 5.00 s, its acceleration is (2.00 m/s2)i hat + (−7.00 m/s2)j. What is the radius of the path taken by the particle if t2 − t1 is less than one period?

Answer 1

Radius of the circular path (R) = 2.95 m

Step-by-step explanation:

Given,

• 
  `t_1\ =\ 2.00 sect_1\ =\ 2.00 sec`
• 
  `\vec{a_1}\ =\ (7.00i\ +\ 2.00j)\ m/s^2`
• 
  `t_2\ =\ 5.0 sec`

• 
  `\vec{a_2}\ =\ (2.0i\ -\ 7.0j)m/s^2`

From the dot product between the two acceleration,

`\therefore \vec{a_1}. \vec{a_2}\ =\ (7.0i\ +\ 2j).(2.0i\ -\ 7.0j)\ =\ 0`

Hence both the accelerations are perpendicular to each other.

Now, given
`t_2\ -\ t_1` is less than the value of the total time period (T).

From the given acceleration the normal components are the same but the radial component of these vectors is opposite in sign, therefore the particle moves a quarter of the circular path or the third part of the circular path. And also the particle is moving in the counterclockwise therefore is will cover the third part of the circular path.

Now the magnitude of the acceleration of the particle is,

`a\ =\ √(7.0^2\ +\ 2.0^2)\ =\ 7.28\ m/s^2`

Let T be the time period of the particle to complete the circular path.

Therefore for the third part of the circular path

`\therefore t_2\ -\ t_1\ =\ (3T)/(4)\\\Rightarrow T\ =\ (4*(5.0\ -\ 2.0))/(3)\\\Rightarrow T\ =\ 4 sec.`

Let R be the radius of the circular motion,

Acceleration of the particle moving in the circular path is

`a\ =\ (4\pi R^2)/(T^2)\\\Rightarrow R\ =\ (aT^2)/(4\pi^2)\\\Rightarrow R\ =\ (7.28* 4^2)/(4* 3.14^2)\\\Rightarrow R\ =\ 2.95\ m`

Hence the radius of the circular motion is 2.95 m.