
A pilot is pulling out of a dive at a speed \[200\;{\rm{m/sec}}\]. If his path is a vertical circle of radius \[1000\;{\rm{meters}}\], then calculate the total force which his body experiences due to the air at the lowest point, if his mass \[60\;{\rm{kg}}\].
Answer
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Hint: The above problem can be resolved using the concepts and applications of linear force and the centripetal force. The linear force acts in either direction that is the force along the straight line. This magnitude of the force is calculated by considering the mass and multiplying it with the magnitude of linear acceleration. On the other hand, the centripetal acceleration is the magnitude of acceleration produced when the body is in a circular motion. Thus, the total force on anybody is the sum of all the forces.
Complete step by step answer:
Given:
The speed of the pilot is, \[v = 200\;{\rm{m/sec}}\].
The radius of the vertical circular path is, \[r = 1000\;{\rm{meters}}\].
The mass of the body is, \[m = 60\;{\rm{kg}}\].
The magnitude of total force is the sum of the linear force acting due to the weight and the magnitude of centripetal force.
The expression is,
\[\begin{array}{l}
F = {F_C} + {F_g}\\
F = \dfrac{{m{v^2}}}{r} + mg\\
F = m\left( {\dfrac{{{v^2}}}{r} + g} \right)
\end{array}\]
Here, g is the gravitational acceleration and its value is \[9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\].
Solve by substituting the values as,
\[\begin{array}{l}
F = m\left( {\dfrac{{{v^2}}}{r} + g} \right)\\
F = 60\;{\rm{kg}} \times \left( {\dfrac{{{{\left( {200\;{\rm{m/sec}}} \right)}^2}}}{{1000\;{\rm{m}}}} + 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\
F = 2988\;{\rm{N}}
\end{array}\]
Therefore, the total force which his body experiences due to the air at the lowest point is of \[2988\;{\rm{N}}\].
Note:To resolve such problems, one must be sure about the concepts and applications of the various types of forces. Out of these forces, there is a wide significance of the centripetal force. The centripetal force is that magnitude of the force that tends to act on the body, changing its angular velocity in the context of time. Moreover, the centripetal force has various applications in resolving the problems of linear as well as rotational kinematics.
Complete step by step answer:
Given:
The speed of the pilot is, \[v = 200\;{\rm{m/sec}}\].
The radius of the vertical circular path is, \[r = 1000\;{\rm{meters}}\].
The mass of the body is, \[m = 60\;{\rm{kg}}\].
The magnitude of total force is the sum of the linear force acting due to the weight and the magnitude of centripetal force.
The expression is,
\[\begin{array}{l}
F = {F_C} + {F_g}\\
F = \dfrac{{m{v^2}}}{r} + mg\\
F = m\left( {\dfrac{{{v^2}}}{r} + g} \right)
\end{array}\]
Here, g is the gravitational acceleration and its value is \[9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\].
Solve by substituting the values as,
\[\begin{array}{l}
F = m\left( {\dfrac{{{v^2}}}{r} + g} \right)\\
F = 60\;{\rm{kg}} \times \left( {\dfrac{{{{\left( {200\;{\rm{m/sec}}} \right)}^2}}}{{1000\;{\rm{m}}}} + 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\
F = 2988\;{\rm{N}}
\end{array}\]
Therefore, the total force which his body experiences due to the air at the lowest point is of \[2988\;{\rm{N}}\].
Note:To resolve such problems, one must be sure about the concepts and applications of the various types of forces. Out of these forces, there is a wide significance of the centripetal force. The centripetal force is that magnitude of the force that tends to act on the body, changing its angular velocity in the context of time. Moreover, the centripetal force has various applications in resolving the problems of linear as well as rotational kinematics.
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