A track consists of two circular parts ABC and CDE of equal radius $100m$ and joined smoothly as shown. Each part subtends a right angle at its centre. A cyclist weighing $100kg$ together with the cycle travels at a constant speed of $18kmph$ on the track. Find the normal force between the road and the cycle just before and just after the cycle crosses C.

A) $682N,682N$
B) $732N,682N$
C) $732N,732N$
D) $682N,732N$
Answer
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Hint: The cyclist undergoes two circular motions. The first one is clockwise and the second one is anticlockwise. There are two forces acting on the cyclist all throughout the motion. The first is the self weight of the cyclist along with the bicycle and the second is the centrifugal force due to the circular motion. The resultant of these forces will help us get the value of normal reaction force.
Complete step by step solution: The original shape of the second law of Newton states that the net force that acts on an object corresponds to its rate of time shift. This law means that if the mass of the object is constant, the acceleration of an object is directly proportional to the net force on the object, in the form of the net force and in inverse proportion to the weight of the object.
Force concepts include the thrust enhancing the object's momentum; the drag decreases the object's velocity; and the torque which changes the object's rotary speed. In an expanded body, the neighbouring sections are typically forceful; the internal mechanical tension is the propagation of such forces through the body.
Radius of the curves $ = 100m$
Weight $ = 100kg$
Velocity $ = 18kmph$
Or, $5m{s^ {- 1}} $
The force acting before the point C,
$mg\cos \theta - N = \dfrac{{m{v^2}}}{R}$
Therefore,
$ \Rightarrow N = mg\cos \theta - \dfrac{{m{v^2}}}{R}$
$ \Rightarrow N = 707 - 25 = 682N$
The force acting between road and cyclist,
\[N - mg\cos \theta = \dfrac{{m{v^2}}}{R}\]
$ \Rightarrow N = mg\cos \theta + \dfrac{{m{v^2}}}{R}$
$ \therefore N = 707 + 25 = 732N$
Therefore, the correct option is D.
Note: The inner mechanical pressures do not allow the body to accelerate while the forces stabilise. Pressure, the movement of several small forces over a body field, is a basic kind of stress that may cause the body to accelerate if the pressure is unbalanced. Stress typically causes rigid material deformation or fluid flow.
Complete step by step solution: The original shape of the second law of Newton states that the net force that acts on an object corresponds to its rate of time shift. This law means that if the mass of the object is constant, the acceleration of an object is directly proportional to the net force on the object, in the form of the net force and in inverse proportion to the weight of the object.
Force concepts include the thrust enhancing the object's momentum; the drag decreases the object's velocity; and the torque which changes the object's rotary speed. In an expanded body, the neighbouring sections are typically forceful; the internal mechanical tension is the propagation of such forces through the body.
Radius of the curves $ = 100m$
Weight $ = 100kg$
Velocity $ = 18kmph$
Or, $5m{s^ {- 1}} $
The force acting before the point C,
$mg\cos \theta - N = \dfrac{{m{v^2}}}{R}$
Therefore,
$ \Rightarrow N = mg\cos \theta - \dfrac{{m{v^2}}}{R}$
$ \Rightarrow N = 707 - 25 = 682N$
The force acting between road and cyclist,
\[N - mg\cos \theta = \dfrac{{m{v^2}}}{R}\]
$ \Rightarrow N = mg\cos \theta + \dfrac{{m{v^2}}}{R}$
$ \therefore N = 707 + 25 = 732N$
Therefore, the correct option is D.
Note: The inner mechanical pressures do not allow the body to accelerate while the forces stabilise. Pressure, the movement of several small forces over a body field, is a basic kind of stress that may cause the body to accelerate if the pressure is unbalanced. Stress typically causes rigid material deformation or fluid flow.
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