Question

A motorcycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be
A. Double
B. Half
C. 4 times
D. \[\dfrac{1}{4}\] times

Answer 1

Hint:Before we start addressing the problem, we need to know about centrifugal force. Centrifugal force is defined as the apparent outward force that acts on a mass when it is rotated. When an object or a particle is moving in a circular motion, then the force which acts away from the center is known as centrifugal force.

Formula Used:
To find the centrifugal force the formula is,
\[F = m{\omega ^2}r\]
Where, m is mass of the body, \[\omega \] is angular velocity and r is radius.

Complete step by step solution:
When a motorcycle driver has a turn, he doubles its velocity, we need to find the force exerted outwards. That is nothing but centrifugal force. The expression for the centrifugal force is,
\[F = m{\omega ^2}r\]
\[\Rightarrow F = \dfrac{{m{v^2}}}{r}\]………… (1)
Now, if the velocity is doubled, that is \[{v^1} = 2v\].
Substitute the value of v in the equation (1) we get,
\[{F^1} = \dfrac{{m{{\left( {2v} \right)}^2}}}{r}\]
\[\Rightarrow {F^1} = \dfrac{{4m{v^2}}}{r}\]
\[\Rightarrow {F^1} = 4\left( {\dfrac{{m{v^2}}}{r}} \right)\]
\[\therefore {F^1} = 4F\]
Therefore, the force exerted outwardly will be 4 times.

Hence, option C is the correct answer.

Note: There is an alternative method to solve this problem which is as shown below.
 When a motorcycle driver has a turn, he doubles its velocity, we need to find the force exerted outwards. That is nothing but centrifugal force. The expression for the centrifugal force is,
\[F = m{\omega ^2}r\]
\[\Rightarrow F = \dfrac{{m{v^2}}}{r}\]………… (1)
Here, \[F \propto {v^2}\]
From this, we can say that the force is directly proportional to the square of the radius. If the velocity doubles, then the force will increase by 4 times. Therefore, the force exerted outwardly will be 4 times.