Photo of rotating scale and falling coins

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

The option (a) says that the 1st coin remains at its earlier position. Due to gravitational force, the 1st coin falls down, hence option (a) is wrong.

I don’t understand the difference between option (B) and (C).

In my opinion, both option says that all of the coins which have left the scale falls down having same position vector.

I am not being able to apply Newton's laws of motion here.

 

[kuruman]

Please complete the template and provide the statement of the problem and the relevant equations.

 

[Pushoam]

Please complete the template and provide the statement of the problem and the relevant equations.

The forces acting on the coin after leaving the scale is m $\vec g$ .

The torque on the scale about the fixed end is

$mg \frac 1 2 L = \frac { mL^2} 3 \alpha$

$\alpha = \frac { 3g } {2L}$

that part of the scale ( which is at a distance L' > $\frac { 2L} 3$ from the pivot ) has an acceleration magnitude $\geq g$ .

So, the coins at a distance L' > $\frac { 2L} 3$ will leave the scale, while the rest of the coin will remain on the scale. Hence, the answer is option (B).

Is this correct?

 

[haruspex]

The option (a) says that the 1st coin remains at its earlier position.

As I read diagrams a to c, they all show the leftmost coins still on the scale, with the rightmost forming a horizontalline below the initial position. They only differ in where the bend in the line is. Yes, diagram a) has the bend very near the left, but not quite.
So you need to solve the problem analytically to find whereabouts the bend shpuld be, then see which diagram looks closest.

I am not being able to apply Newton's laws of motion here.

Why not? Try to determine the initial angular acceleration of the scale.

 

[Pushoam]

Why not? Try to determine the initial angular acceleration of the scale.

Please see the post # 2.

 

[haruspex]

The forces acting on the coin after leaving the scale is m $\vec g$ .

The torque on the scale about the fixed end is

$mg \frac 1 2 L = \frac { mL^2} 3 \alpha$

$\alpha = \frac { 3g } {2L}$

that part of the scale ( which is at a distance L' > $\frac { 2L} 3$ from the pivot ) has an acceleration magnitude $\geq g$ .

So, the coins at a distance L' > $\frac { 2L} 3$ will leave the scale, while the rest of the coin will remain on the scale. Hence, the answer is option (B).

Is this correct?

Looks good.