Answer
Verified
453.3k+ views
Hint
By drawing the free body diagram of the body at the lowest point, we can find the equation for the body due to its weight, the centrifugal force and the normal reaction acting on the body due to the road. We can find the thrust as it is equal to the normal reaction force on the body due to the road.
Formula Used: In this solution we will be using the following formula,
$ {F_c} = \dfrac{{m{v^2}}}{r} $
where $ {F_c} $ is the centrifugal force,
$ m $ is the mass of the body
$ v $ is the velocity of the body
$ r $ is the radius of the concave path.
Complete step by step answer
Let us first draw the image of a body at the lowest point on a concave road with the forces acting on it.
Now as we can see that for the body, the weight will be acting in the downward direction and the normal reaction force acting on the body will be in the upward direction. The velocity of the body will be as given in the diagram as a tangent to the circular path.
Now let us consider the radius of this circular path to be $ r $ . Now since the body is moving in a circular path, the force that is acting on the body is called centrifugal force. This force acts in a direction which is outward on the circular path. So the centrifugal force will be acting on the body in the upward direction.
So the sum of the centrifugal force on the body and the mass of the body will be balanced by the normal reaction force on the body due to the road.
Therefore, we can write
$ N = mg + {F_c} $
The centrifugal acceleration is given as,
$ {F_c} = \dfrac{{m{v^2}}}{r} $
So substituting this in the equation we get,
$ N = mg + \dfrac{{m{v^2}}}{r} $
Since this normal reaction force is the thrust on the road, so the answer will be, $ mg + \dfrac{{m{v^2}}}{r} $
Hence the correct answer is option (B).
Note
If the body was moving on a convex road, then the centrifugal force on the body due to the curved path would be acting in the upward direction. Therefore, the thrust on the road will be the weight on the body minus the centrifugal force.
By drawing the free body diagram of the body at the lowest point, we can find the equation for the body due to its weight, the centrifugal force and the normal reaction acting on the body due to the road. We can find the thrust as it is equal to the normal reaction force on the body due to the road.
Formula Used: In this solution we will be using the following formula,
$ {F_c} = \dfrac{{m{v^2}}}{r} $
where $ {F_c} $ is the centrifugal force,
$ m $ is the mass of the body
$ v $ is the velocity of the body
$ r $ is the radius of the concave path.
Complete step by step answer
Let us first draw the image of a body at the lowest point on a concave road with the forces acting on it.
Now as we can see that for the body, the weight will be acting in the downward direction and the normal reaction force acting on the body will be in the upward direction. The velocity of the body will be as given in the diagram as a tangent to the circular path.
Now let us consider the radius of this circular path to be $ r $ . Now since the body is moving in a circular path, the force that is acting on the body is called centrifugal force. This force acts in a direction which is outward on the circular path. So the centrifugal force will be acting on the body in the upward direction.
So the sum of the centrifugal force on the body and the mass of the body will be balanced by the normal reaction force on the body due to the road.
Therefore, we can write
$ N = mg + {F_c} $
The centrifugal acceleration is given as,
$ {F_c} = \dfrac{{m{v^2}}}{r} $
So substituting this in the equation we get,
$ N = mg + \dfrac{{m{v^2}}}{r} $
Since this normal reaction force is the thrust on the road, so the answer will be, $ mg + \dfrac{{m{v^2}}}{r} $
Hence the correct answer is option (B).
Note
If the body was moving on a convex road, then the centrifugal force on the body due to the curved path would be acting in the upward direction. Therefore, the thrust on the road will be the weight on the body minus the centrifugal force.
Recently Updated Pages
A wire of length L and radius r is clamped rigidly class 11 physics JEE_Main
For which of the following reactions H is equal to class 11 chemistry JEE_Main
For the redox reaction MnO4 + C2O42 + H + to Mn2 + class 11 chemistry JEE_Main
In the reaction 2FeCl3 + H2S to 2FeCl2 + 2HCl + S class 11 chemistry JEE_Main
One mole of a nonideal gas undergoes a change of state class 11 chemistry JEE_Main
A stone is projected with speed 20 ms at angle 37circ class 11 physics JEE_Main
Trending doubts
Which is the longest day and shortest night in the class 11 sst CBSE
Who was the Governor general of India at the time of class 11 social science CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Define the term system surroundings open system closed class 11 chemistry CBSE
In a democracy the final decisionmaking power rests class 11 social science CBSE
What organs are located on the left side of your body class 11 biology CBSE