
A 50 kg passenger on an amusement park ride stands with his back against the wall of a cylindrical room with a radius of 3 m. What is the centripetal force of the wall pressing in to his back, when the room spins and he is moving at 6m/s
${\text{A) 60N}}$
${\text{B) 600N}}$
${\text{C) 36N}}$
${\text{D) 360N}}$
Answer
596.4k+ views
Hint: For any body to continue in a circular motion, it needs force acting towards the center known as centripetal force. Centripetal force is not a fundamental force. It is the net force which makes the object to move in a circular path. Even multiple forces can be involved to produce a net force.
Formula used:
${F_c} = \dfrac{{m{v^2}}}{R}$,
${F_c}$ is the centripetal force, $m$ is the mass of the object, $v$ is the velocity of the object and $R$ is the radius of the circular path.
Complete step by step answer:
According to Newton’s first law, the external force needed to change the direction in a circular motion is provided by the centripetal force. Centripetal acceleration is given by ${a_c} = \dfrac{{{v^2}}}{R}$.
Using Newton’s second law,
$\dfrac{{{v^2}}}{R} = \dfrac{F}{m}$
$\therefore {F_c} = \dfrac{{m{v^2}}}{R}$
${F_c} = \dfrac{{m{v^2}}}{R}$
${F_c}$ is the centripetal force, $m$ is the mass of the object,$v$ is the velocity of the object and $R$ is the radius of the circular path.
Given $m = 50kg$, $v = 6m/s$ and $R = 3m$.Using ${F_c} = \dfrac{{m{v^2}}}{R}$
${F_c} = \dfrac{{50 \times {6^2}}}{3}$
$\Rightarrow {F_c} = \dfrac{{50 \times 36}}{3}$
$\Rightarrow {F_c} = 50 \times 12$
$\Rightarrow {F_c} = 600N$
The correct option is (B)
Additional Information
Newton’s First Law: Any body continues to be in a state of rest or uniform motion until an external unbalanced force acts on it.
Newton’s second law: In an inertial frame of reference the mass multiplied by the acceleration of the object gives the vector sum of the forces $F$.
Note: Another force related to circular motion is centrifugal force. Centripetal force is an actual force whereas centrifugal force is an apparent force. Centrifugal Force is the apparent force that is felt by an object moving in a curved path that acts outwardly away from the center of motion.
Formula used:
${F_c} = \dfrac{{m{v^2}}}{R}$,
${F_c}$ is the centripetal force, $m$ is the mass of the object, $v$ is the velocity of the object and $R$ is the radius of the circular path.
Complete step by step answer:
According to Newton’s first law, the external force needed to change the direction in a circular motion is provided by the centripetal force. Centripetal acceleration is given by ${a_c} = \dfrac{{{v^2}}}{R}$.
Using Newton’s second law,
$\dfrac{{{v^2}}}{R} = \dfrac{F}{m}$
$\therefore {F_c} = \dfrac{{m{v^2}}}{R}$
${F_c} = \dfrac{{m{v^2}}}{R}$
${F_c}$ is the centripetal force, $m$ is the mass of the object,$v$ is the velocity of the object and $R$ is the radius of the circular path.
Given $m = 50kg$, $v = 6m/s$ and $R = 3m$.Using ${F_c} = \dfrac{{m{v^2}}}{R}$
${F_c} = \dfrac{{50 \times {6^2}}}{3}$
$\Rightarrow {F_c} = \dfrac{{50 \times 36}}{3}$
$\Rightarrow {F_c} = 50 \times 12$
$\Rightarrow {F_c} = 600N$
The correct option is (B)
Additional Information
Newton’s First Law: Any body continues to be in a state of rest or uniform motion until an external unbalanced force acts on it.
Newton’s second law: In an inertial frame of reference the mass multiplied by the acceleration of the object gives the vector sum of the forces $F$.
Note: Another force related to circular motion is centrifugal force. Centripetal force is an actual force whereas centrifugal force is an apparent force. Centrifugal Force is the apparent force that is felt by an object moving in a curved path that acts outwardly away from the center of motion.
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