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

[Karol]

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:
$$\frac{1}{2}mg(L-D)+2\cdot mg(L-D)=mgL\;\rightarrow\; D=\frac{3}{5}L$$

 

[ehild]

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.

 

[Karol]

Thank you Ehild