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Answer:
1.23 m
Explanation:
The vertical distance covered by a free-falling object starting from rest in a time t is
where
g = 9.8 m/s^2 is the acceleration due to gravity
In this problem, we have
t = 0.50 s
So the distance covered is
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;
<em>1: m = 20 kg, r = 0.25·R</em>
<em>2: m = 10 kg, r = 1.0·R</em>
<em>3: m = 10 kg, r = 0.25·R</em>
<em>4: m = 15 kg, r = 0.75·R</em>
<em>5: m = 10 kg, r = 0.5·R</em>
<em>6: m = 40 kg, r = 0.25·R</em>
According to the principle of conservation of angular momentum, we have;
The moment of inertia of the merry-go-round, = 0.5·M·R²
Moment of inertia of the sandbag = m·r²
Therefore;
0.5·M·R²· = (0.5·M·R² + m·r²)·
Given that 0.5·M·R²· is constant, as the value of m·r² increases, the value of
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;
[<u>m = 10 kg, r = 0.25·R</u>] > [<u>m = 20 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 0.5·R</u>] > [<u>m = </u>
<u>10 kg, r = 0.5·R</u>] = [<u>m = 40 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 1.0·R</u>].
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