Question

A man whose mental health can be seriously questioned is trying to bicycle through an ellipse-shaped loop. The total mass of the man and his bicycle is 85 kg and the center of mass is 1.2 m from the ground. What is the minimum speed he has to have at A in order to be able to perform this stunt without embarrassing (and potentially lethal) results? Neglect friction, air resistance and the size of wheels. 1.2 m G y B 3 m ² + ² = 1 4 m

Answer 1

In order for the man to safely complete the loop-the-loop stunt without falling due to gravity, his centripetal force at the top of the loop (point B) should be greater than or equal to the gravitational force pulling him downward. The centripetal force required at the top of the loop is given by:

${�}_{�}=\frac{�{�}^2}{�}$,

Where:

• ${�}_{�}$ is the centripetal force required at the top of the loop.

• $�$ is the total mass of the man and his bicycle, which is 85 kg.

• $�$ is the minimum speed he needs to have at point A.

• $�$ is the radius of the loop, which is the distance from the center of mass to the top of the loop (3 m + 1.2 m = 4.2 m).

The gravitational force pulling the man and his bicycle downward is given by:

${�}_{�}=��$,

Where:

• ${�}_{�}$ is the gravitational force.

• $�$ is the total mass of the man and his bicycle, which is 85 kg.

• $�$ is the acceleration due to gravity, which is approximately $9.81�\mathrm{/}{�}^2$.

Now, we can set up the condition for a successful loop-the-loop:

${�}_{�}\ge {�}_{�}$.

$\frac{�{�}^2}{�}\ge ��$.

Now, plug in the values:

$\frac{85��\cdot {�}^2}{4.2�}\ge 85��\cdot 9.81�\mathrm{/}{�}^2$.

Simplify:

${�}^2\ge 4.2\cdot 9.81$.

${�}^2\ge 41.202$.

Now, take the square root of both sides:

$�\ge \sqrt{41.202}�\mathrm{/}�$.

Calculating this value:

$�\ge 6.42�\mathrm{/}�$.

So, the minimum speed he needs to have at point A is approximately 6.42 meters per second to successfully complete the loop-the-loop without falling due to gravity.

Answer 2

In order for the man to safely complete the loop-the-loop stunt without falling due to gravity, his centripetal force at the top of the loop (point B) should be greater than or equal to the gravitational force pulling him downward. The centripetal force required at the top of the loop is given by:

${�}_{�}=\frac{�{�}^2}{�}$,

Where:

• ${�}_{�}$ is the centripetal force required at the top of the loop.

• $�$ is the total mass of the man and his bicycle, which is 85 kg.

• $�$ is the minimum speed he needs to have at point A.

• $�$ is the radius of the loop, which is the distance from the center of mass to the top of the loop (3 m + 1.2 m = 4.2 m).

The gravitational force pulling the man and his bicycle downward is given by:

${�}_{�}=��$,

Where:

• ${�}_{�}$ is the gravitational force.

• $�$ is the total mass of the man and his bicycle, which is 85 kg.

• $�$ is the acceleration due to gravity, which is approximately $9.81�\mathrm{/}{�}^2$.

Now, we can set up the condition for a successful loop-the-loop:

${�}_{�}\ge {�}_{�}$.

$\frac{�{�}^2}{�}\ge ��$.

Now, plug in the values:

$\frac{85��\cdot {�}^2}{4.2�}\ge 85��\cdot 9.81�\mathrm{/}{�}^2$.

Simplify:

${�}^2\ge 4.2\cdot 9.81$.

${�}^2\ge 41.202$.

Now, take the square root of both sides:

$�\ge \sqrt{41.202}�\mathrm{/}�$.

Calculating this value:

$�\ge 6.42�\mathrm{/}�$.

So, the minimum speed he needs to have at point A is approximately 6.42 meters per second to successfully complete the loop-the-loop without falling due to gravity.