Question

The figure shows the velocity and acceleration of a point-like body at the initial moment of its motion. The acceleration vector of the body remains constant. The minimum radius of curvature of trajectory of the body is

A. 2 meters
B. 4 meters
C. 8 meters
D. 16 meters

Answer 1

In the question it is asked what will be the minimum radius of curvature of trajectory of the body. This means the body should perform a circular motion for that and for a body to perform the circular motion the force acting on the body should be perpendicular to the velocity of the body. So, we will find the perpendicular velocity of the point mass.

Complete step-by-step answer:
Given,
There is a point body which is having-
Initial velocity, ${v_0} = 8$ m/s
Acceleration, $a = 2$ m/s2 (it is given in the question that the acceleration is constant)
Angle between acceleration and velocity, $\theta = 150^\circ$
In the given diagram we will do some construction, that is we will extend the acceleration vector in the opposite direction and we will take components of the velocity vector.

So, we have done the components of the velocity vector as above. We also know that the direction of the force will be in the same direction as that of the acceleration, so the horizontal component of the velocity will not contribute in the circular motion of the point object (since the circular motion is required to get the curvature of trajectory). So only the perpendicular component which is ${v_0}\cos \theta$ will be responsible for the circular motion of the point mass.
As now we know that the point mass will perform the circular motion then the acceleration which is given will act like centripetal acceleration.
And the centripetal acceleration of a point mass is given by,
${a_c} = \dfrac{{{v_p}^2}}{R}$ , where ${a_c} =$ centripetal acceleration, $R =$ radius of curvature,
${v_p} =$ perpendicular component of the velocity.
Here we have ${v_p} = {v_0}\sin \theta$
And ${a_c} = 2$ m/s2
So, the radius of curvature,
$R = \dfrac{{{v_p}^2}}{{{a_c}}}$
Now putting the values,
$R = \dfrac{{{{({v_0}\sin \theta )}^2}}}{2}$
$\Rightarrow R = \dfrac{{{{(8 \times \dfrac{1}{2})}^2}}}{2}$
$\Rightarrow R = \dfrac{{16}}{2}$
$\Rightarrow R = 8$ metre.
Hence the minimum radius of curvature of trajectory of the body is $8$ metre.

So, the correct answer is “Option C”.

Note: In a circular motion we know that there should be a force which is responsible for the circular motion of the particle or body. That particular force is known as the centripetal force. The direction of the centripetal force is always towards the centre and also the direction of the acceleration is also towards the centre of the trajectory. So, this implies that in a circular motion the direction of the centripetal force and the direction of the centripetal acceleration is in the same direction.