Question

A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so small that it can be ignored. A sandbag of mass m is dropped onto the merry-go-round, at a position designated by r. The sandbag does not slip or roll upon contact with the merry-go-round.
The figure shows a top view of a merry-go-round of radius capital R rotating counterclockwise. A sandbag is located on the merry-go-round at a distance lowercase r from the center.

Rank the following different combinations of m and r on the basis of the angular speed of the merry-go-round after the sandbag "sticks" to the merry-go-round.

Answer 1

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              ${}$ ⇩

[m = 20 kg, r = 0.25·R]

              ${}$ ⇩

[m = 10 kg, r = 0.5·R]

              ${}$ ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              ${}$ ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

1: m = 20 kg, r = 0.25·R

2: m = 10 kg, r = 1.0·R

3: m = 10 kg, r = 0.25·R

4: m = 15 kg, r = 0.75·R

5: m = 10 kg, r = 0.5·R

6: m = 40 kg, r = 0.25·R

According to the principle of conservation of angular momentum, we have;

$I_i \cdot \omega _i = I_f \cdot \omega _f$

The moment of inertia of the merry-go-round, $I_m$ = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·$\omega _i$ = (0.5·M·R² + m·r²)·$\omega _f$

Given that 0.5·M·R²·$\omega _i$ is constant, as the value of  m·r² increases, the value of $\omega _f$ decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =

10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].

Learn more here:

brainly.com/question/15188750