
A gymnast hangs from one end of a rope and executes vertical circular motion. The gymnast has a mass of $40{\text{ }}Kg$ and the radius of the circle is $2.5{\text{ m}}$. If the gymnast whirls himself at a constant speed of $6{\text{ m}}{{\text{s}}^{ - 1}}$, what is the tension in the rope at the lowest point? (in N)
Answer
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Hint: in order to solve the question, we will use the formula centripetal force which is applied in the executes vertical circular motion and we will use the gravitational force which is applied because of the mass of gymnast and the gravity which is applied on him to find the force which is applied against the tension in the rope at the lowest point.
Formula used:
${F_c} = \dfrac{{m{v^2}}}{r}$
Here, $F$ refers to centripetal force, $m$ refers to mass, $v$ refers to velocity and $r$ refers to radius.
${F_g} = mg$
$F$ refers to force by gravity, $m$ refers to mass and $g$ refers to acceleration of gravity.
Complete step by step answer:
In the question we are given that there is a gymnast who hangs from one end of a rope and executes vertical circular motion.
Gymnast has a mass of,
$m = 40{\text{ }}Kg$
Radius of the circle is,
$r = 2.5{\text{ m}}$.
gymnast whirls himself at a constant speed
$v = 6{\text{ m}}{{\text{s}}^{ - 1}}$
And we have to find the tension in the rope at the lowest point.
The tension in the rope at the lowest point can be find by applying the centripetal force and force of gravitation
${F_c} = \dfrac{{m{v^2}}}{r}$
$\Rightarrow {F_g} = mg$
We have to added the centripetal force and force of gravitation and we will get tension as both the force pull at the lowest point which makes it equal to tension
$T = {F_c} + {F_g}$
Expanding the formula using the equations above
$T = \dfrac{{m{v^2}}}{r} + mg$
Now putting the values of m, v, r, g
$T = \dfrac{{40 kg \times 6 m{s^{ - 1}}}}{{2.5 m}} + 40 kg \times 9.8 m{s^{ - 2}}$
After doing the calculation we get tension equals to
$\therefore T = 968{\text{ N}}$
Hence, the correct answer is \[968{\text{ N}}\] tension in the rope at the lowest point.
Note: Many of the people will make the mistake by not considering the gravitational force in action and only apply the centripetal force because of vertical motion but until it is not mention in question we have to consider gravity and we don’t have to consider air resistance in the way of gymnast until mention in the question.
Formula used:
${F_c} = \dfrac{{m{v^2}}}{r}$
Here, $F$ refers to centripetal force, $m$ refers to mass, $v$ refers to velocity and $r$ refers to radius.
${F_g} = mg$
$F$ refers to force by gravity, $m$ refers to mass and $g$ refers to acceleration of gravity.
Complete step by step answer:
In the question we are given that there is a gymnast who hangs from one end of a rope and executes vertical circular motion.
Gymnast has a mass of,
$m = 40{\text{ }}Kg$
Radius of the circle is,
$r = 2.5{\text{ m}}$.
gymnast whirls himself at a constant speed
$v = 6{\text{ m}}{{\text{s}}^{ - 1}}$
And we have to find the tension in the rope at the lowest point.
The tension in the rope at the lowest point can be find by applying the centripetal force and force of gravitation
${F_c} = \dfrac{{m{v^2}}}{r}$
$\Rightarrow {F_g} = mg$
We have to added the centripetal force and force of gravitation and we will get tension as both the force pull at the lowest point which makes it equal to tension
$T = {F_c} + {F_g}$
Expanding the formula using the equations above
$T = \dfrac{{m{v^2}}}{r} + mg$
Now putting the values of m, v, r, g
$T = \dfrac{{40 kg \times 6 m{s^{ - 1}}}}{{2.5 m}} + 40 kg \times 9.8 m{s^{ - 2}}$
After doing the calculation we get tension equals to
$\therefore T = 968{\text{ N}}$
Hence, the correct answer is \[968{\text{ N}}\] tension in the rope at the lowest point.
Note: Many of the people will make the mistake by not considering the gravitational force in action and only apply the centripetal force because of vertical motion but until it is not mention in question we have to consider gravity and we don’t have to consider air resistance in the way of gymnast until mention in the question.
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