
A car of mass 1000 kg moves on a circular track of radius 40 m. If the coefficient of friction is 1.28 then the maximum velocity with which the car can be moved is:
A.\[22.4m{{s}^{-1}}\]
B.\[112m{{s}^{-1}}\]
C.\[v=\sqrt{1.28\times 10\times 40}=22.4m{{s}^{-1}}\]
D.\[1000m{{s}^{-1}}\]
Answer
586.2k+ views
Hint: Here the centripetal acceleration for the motion in circular path is given by the frictional force acting on the car. Centripetal acceleration is the characteristics in the movement of a body in circular motion. The body is attracted towards the centre by centripetal force and the centripetal acceleration is radially towards the centre.
Formula used:
\[m\dfrac{{{v}^{2}}}{r}=\mu mg\]
Complete step-by-step answer:
First of all let us look at the definition of centripetal acceleration and its properties. Centripetal acceleration is the characteristics in the movement of a body in circular motion. The body is attracted towards the centre by centripetal force and the centripetal acceleration is radially towards the centre. It is having magnitude which is equal to the square of speed of the body which is further divided with the radius of circular path covered.
Centripetal acceleration is given by the equation
\[a=\dfrac{{{v}^{2}}}{r}\]
Here v is the velocity of the body and r is the distance covered radially.
Next let us talk about the frictional force. It is an opposing force that acts on a moving body to prevent its motion. Frictional force is the reason why a body kept on a table is at rest until any external force greater than frictional force acts on it.
The equation of frictional force is
\[F=\mu mg\]
So here the centripetal acceleration is provided by the frictional force acting there.
So as per the conditions, we can write that,
\[m\dfrac{{{v}^{2}}}{r}=\mu mg\]
Left hand side and right hand side are having mass, so we can cancel it.
Therefore
\[v=\sqrt{\mu gr}\]
Substituting the values in the equation,
\[v=\sqrt{1.28\times 10\times 40}=22.4m{{s}^{-1}}\]
Therefore the correct answer is option A.
Note: The concept of equalising the centripetal force to the frictional force is the key concept here. The formula of centripetal force and frictional force should be known. That means
\[m\dfrac{{{v}^{2}}}{r}=\mu mg\]
The frictional force is an opposing force that acts on a moving body to prevent its motion.
Formula used:
\[m\dfrac{{{v}^{2}}}{r}=\mu mg\]
Complete step-by-step answer:
First of all let us look at the definition of centripetal acceleration and its properties. Centripetal acceleration is the characteristics in the movement of a body in circular motion. The body is attracted towards the centre by centripetal force and the centripetal acceleration is radially towards the centre. It is having magnitude which is equal to the square of speed of the body which is further divided with the radius of circular path covered.
Centripetal acceleration is given by the equation
\[a=\dfrac{{{v}^{2}}}{r}\]
Here v is the velocity of the body and r is the distance covered radially.
Next let us talk about the frictional force. It is an opposing force that acts on a moving body to prevent its motion. Frictional force is the reason why a body kept on a table is at rest until any external force greater than frictional force acts on it.
The equation of frictional force is
\[F=\mu mg\]
So here the centripetal acceleration is provided by the frictional force acting there.
So as per the conditions, we can write that,
\[m\dfrac{{{v}^{2}}}{r}=\mu mg\]
Left hand side and right hand side are having mass, so we can cancel it.
Therefore
\[v=\sqrt{\mu gr}\]
Substituting the values in the equation,
\[v=\sqrt{1.28\times 10\times 40}=22.4m{{s}^{-1}}\]
Therefore the correct answer is option A.
Note: The concept of equalising the centripetal force to the frictional force is the key concept here. The formula of centripetal force and frictional force should be known. That means
\[m\dfrac{{{v}^{2}}}{r}=\mu mg\]
The frictional force is an opposing force that acts on a moving body to prevent its motion.
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