# Mass attached to a rope. the rope encounters a nail.

• Karol
In summary, the conversation discusses the motion of an object attached to a rope and hitting a nail at a certain distance. Using the equations for kinetic and potential energy, as well as centrifugal force, it is determined that the object's velocity at the lowest point is equal to the square root of 2 times the gravitational acceleration times the length of the rope. The minimal distance for the nail to be placed in order for the object to complete a circle around it is determined to be 3/5 of the length of the rope.

## Homework Statement

M is attached to a rope L and starts from horizontal. the rope hits a nail at distance D.
What is m's velocity when it reaches the lowest point.
At what minimal distance D should the nail be places so that m will complete a circle round the nail.

## Homework Equations

Kinetic energy: ##E_k=\frac{1}{2}mv^2##
Potential energy: ##E_p=mgh##
Centrifugal force: ##F=m\frac{v^2}{r}##

## The Attempt at a Solution

Conservation of energy relative to the lowest point:
$$mgL=\frac{1}{2}mv^2\;\rightarrow\; v=\sqrt{2gL}$$
The centrifugal force at the top equals only the weight:
$$m\frac{v_2^2}{L-D}=mg\;\rightarrow\;v_2^2=(L-D)g$$
The kinetic energy at the bottom, which equals the initial potential energy, equals the potential and kinetic energy at the top:
$$\frac{1}{2}mg(L-D)+2\cdot mg(L-D)=mgL\;\rightarrow\; D=\frac{3}{5}L$$

Karol said:

## Homework Statement

M is attached to a rope L and starts from horizontal. the rope hits a nail at distance D.
What is m's velocity when it reaches the lowest point.
At what minimal distance D should the nail be places so that m will complete a circle round the nail.
View attachment 99213

## Homework Equations

Kinetic energy: ##E_k=\frac{1}{2}mv^2##
Potential energy: ##E_p=mgh##
Centrifugal force: ##F=m\frac{v^2}{r}##

## The Attempt at a Solution

Conservation of energy relative to the lowest point:
$$mgL=\frac{1}{2}mv^2\;\rightarrow\; v=\sqrt{2gL}$$
The centrifugal force at the top equals only the weight:
$$m\frac{v_2^2}{L-D}=mg\;\rightarrow\;v_2^2=(L-D)g$$
The kinetic energy at the bottom, which equals the initial potential energy, equals the potential and kinetic energy at the top:
$$\frac{1}{2}mg(L-D)+2\cdot mg(L-D)=mgL\;\rightarrow\; D=\frac{3}{5}L$$
It looks correct.

Thank you Ehild

## 1. How does the mass of the object affect the tension in the rope?

The mass of the object attached to the rope will determine the amount of tension in the rope. The larger the mass, the greater the tension will be.

## 2. What happens to the velocity of the mass when it encounters the nail?

When the mass attached to a rope encounters a nail, its velocity will decrease as it experiences a sudden change in direction and may even come to a complete stop depending on the strength of the nail and the force of the impact.

## 3. Does the length of the rope have an effect on the force applied to the nail?

Yes, the length of the rope can affect the force applied to the nail. A longer rope will have more slack and will not be able to apply as much force as a shorter rope with less slack.

## 4. Can the rope break when encountering a nail?

It is possible for the rope to break when encountering a nail, especially if the force applied to the nail is greater than the tensile strength of the rope. This can also depend on the material and thickness of the rope.

## 5. How does the angle of the rope affect the force applied to the nail?

The angle of the rope can affect the force applied to the nail. A rope at a steeper angle will apply more force to the nail compared to a rope at a shallower angle. This is due to the component of the force acting in the direction of the nail being greater at a steeper angle.