
A fly over bridge is a part of a vertical circle of radius 80m and a vehicle is travelling on it. The vehicle does not leave the surface of the bridge at its highest point. If maximum speed of vehicle is
A. $14\,m/s$
B. $21\,m/s$
C. $28\,m/s$
D. $42\,m/s$
Answer
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Hint: At the highest point when the velocity is maximum the only force acting on the vehicle is the force due to gravity, that is weight. We know that the force providing circular motion is the centripetal force. So, weight is providing the necessary centripetal force here. Using this we can find the equation for finding maximum velocity.
Complete step by step answer:
It is given that a flyover bridge is part of a vertical circle of radius $80\,m$ and a vehicle is travelling on it.
$ \Rightarrow r = 80\,m$
We need to find the maximum speed of the vehicle so that it does not leave the surface of the bridge at its highest point.
we know that in a circular motion the force providing the circular motion is the centripetal force.
As velocity increases the centripetal force increases and at the maximum height the normal reaction can be taken as zero. Hence the only force acting there is the weight. This force due to weight is that what provides the necessary centripetal force. So, let us equate the weight with the centripetal force.
Centripetal force is given as
${F_c} = \dfrac{{m{v^2}}}{r}$
Where, m is the mass, v is the velocity and r is the radius of the circular path.
Force of weight is given as $W = m\,g$
Where, g is the acceleration due to gravity.
$g = 9.8\,m/s$
On equating both the forces, we get
$ \Rightarrow \dfrac{{m{v^2}}}{r} = mg$
$ \Rightarrow \dfrac{{{v^2}}}{r} = g$
$ \Rightarrow v = \sqrt {gr} $
This the critical velocity or the maximum velocity. If velocity is greater than this, then the vehicle will not follow a circular path it will fly off.
Now let us substitute the given values in the equation for maximum velocity. Then we get
$ \Rightarrow v = \sqrt {gr} $
$ \Rightarrow v = \sqrt {9.8 \times 80} $
$ \Rightarrow v = 28\,m/s$
This is the value of maximum speed of the vehicle.
Hence the correct answer is option C.
Note:
Remember that at the highest point, when the velocity is maximum the normal reaction will be zero. The gravitational force will be the only force providing the necessary centripetal force at this point. On any other point on the circular path the net force between the weight and the normal reaction is providing the centripetal acceleration.
Complete step by step answer:
It is given that a flyover bridge is part of a vertical circle of radius $80\,m$ and a vehicle is travelling on it.
$ \Rightarrow r = 80\,m$
We need to find the maximum speed of the vehicle so that it does not leave the surface of the bridge at its highest point.
we know that in a circular motion the force providing the circular motion is the centripetal force.
As velocity increases the centripetal force increases and at the maximum height the normal reaction can be taken as zero. Hence the only force acting there is the weight. This force due to weight is that what provides the necessary centripetal force. So, let us equate the weight with the centripetal force.
Centripetal force is given as
${F_c} = \dfrac{{m{v^2}}}{r}$
Where, m is the mass, v is the velocity and r is the radius of the circular path.
Force of weight is given as $W = m\,g$
Where, g is the acceleration due to gravity.
$g = 9.8\,m/s$
On equating both the forces, we get
$ \Rightarrow \dfrac{{m{v^2}}}{r} = mg$
$ \Rightarrow \dfrac{{{v^2}}}{r} = g$
$ \Rightarrow v = \sqrt {gr} $
This the critical velocity or the maximum velocity. If velocity is greater than this, then the vehicle will not follow a circular path it will fly off.
Now let us substitute the given values in the equation for maximum velocity. Then we get
$ \Rightarrow v = \sqrt {gr} $
$ \Rightarrow v = \sqrt {9.8 \times 80} $
$ \Rightarrow v = 28\,m/s$
This is the value of maximum speed of the vehicle.
Hence the correct answer is option C.
Note:
Remember that at the highest point, when the velocity is maximum the normal reaction will be zero. The gravitational force will be the only force providing the necessary centripetal force at this point. On any other point on the circular path the net force between the weight and the normal reaction is providing the centripetal acceleration.
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