
A 100 kg car is moving with a maximum velocity of 9 m/s across a circular track of radius 30 m. The maximum force of friction between the road and the car is
A. 1000 N
B. 706 N
C. 270 N
D. 200N
Answer
161.7k+ views
Hint: When the car is moving on a circular track then an outward centrifugal force acts on the body which is proportional to the square of the linear velocity of the car. To keep the car on the circular track without skidding there is a balancing force inward which is a static frictional force.
Formula used:
\[{F_0} = \dfrac{{m{v^2}}}{r}\]
Here \[{F_0}\] is the outward force acting on the body of mass m in a circular path of radius r moving with linear velocity v.
\[{f_s} = \mu N\]
Here \[{f_s}\] is the static friction between the tyre of the car and the road with a coefficient of friction \[\mu \] and N is the normal force acting.
Complete step by step solution:
The normal force acting on the car due to the surface of the road is equal to the weight of the car,
\[N = mg\]
The outward force acting on the car in the circular track is,
\[{F_o} = \dfrac{{mv_{\max }^2}}{r}\]
Here, \[{v_{\max }}\] is the maximum velocity for the skidding of the car.
The static friction between the car and the road is,
\[{f_s} = \mu mg\]
At the equilibrium condition for the car along the line joining the center of the circular track and the car, the net force along the line must be zero, i.e. the inward static frictional force must balance the outward centrifugal force. Hence, the magnitude of these two forces must be equal.
\[{F_o} = {f_s}\]
So, the maximum kinetic energy acting between the tire of the car and the road is equal to the centrifugal force.
\[{f_s} = \dfrac{{mv_{\max }^2}}{r}\]
The mass of the car is 100 kg and the speed is 9 m/s. The radius of the circular track is given as 30 m. Putting the values, we get
\[{f_s} = \dfrac{{100 \times {{\left( 9 \right)}^2}}}{{30}}N\]
\[\Rightarrow {f_s} = \dfrac{{100 \times 81}}{{30}}N\]
\[\therefore {f_s} = 270\,N\]
Therefore, the correct option is C.
Note: If we increase the speed of the car more than the maximum velocity for the skidding then the outward force on the car will be greater than the static friction acting on it, and its result in the skidding of the car outside the circular track.
Formula used:
\[{F_0} = \dfrac{{m{v^2}}}{r}\]
Here \[{F_0}\] is the outward force acting on the body of mass m in a circular path of radius r moving with linear velocity v.
\[{f_s} = \mu N\]
Here \[{f_s}\] is the static friction between the tyre of the car and the road with a coefficient of friction \[\mu \] and N is the normal force acting.
Complete step by step solution:
The normal force acting on the car due to the surface of the road is equal to the weight of the car,
\[N = mg\]
The outward force acting on the car in the circular track is,
\[{F_o} = \dfrac{{mv_{\max }^2}}{r}\]
Here, \[{v_{\max }}\] is the maximum velocity for the skidding of the car.
The static friction between the car and the road is,
\[{f_s} = \mu mg\]
At the equilibrium condition for the car along the line joining the center of the circular track and the car, the net force along the line must be zero, i.e. the inward static frictional force must balance the outward centrifugal force. Hence, the magnitude of these two forces must be equal.
\[{F_o} = {f_s}\]
So, the maximum kinetic energy acting between the tire of the car and the road is equal to the centrifugal force.
\[{f_s} = \dfrac{{mv_{\max }^2}}{r}\]
The mass of the car is 100 kg and the speed is 9 m/s. The radius of the circular track is given as 30 m. Putting the values, we get
\[{f_s} = \dfrac{{100 \times {{\left( 9 \right)}^2}}}{{30}}N\]
\[\Rightarrow {f_s} = \dfrac{{100 \times 81}}{{30}}N\]
\[\therefore {f_s} = 270\,N\]
Therefore, the correct option is C.
Note: If we increase the speed of the car more than the maximum velocity for the skidding then the outward force on the car will be greater than the static friction acting on it, and its result in the skidding of the car outside the circular track.
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