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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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Answer:
The magnitude of the force that each wire exerts on the other will increase by a factor of two.
Explanation:
force on parallel current carrying wire, F = BILsinθ
where;
B is the strength of the magnetic field
L is the length of the wire
I is the magnitude of current on the wire
θ is the angle of inclination of the wire
Assuming B, L and θ is constant, then F ∝ I
F = kI
When the amount of current is doubled in one of the wires, lets say the second wire;
Also, if will double the amount of current on the first wire, then
F₁ = 2F₂
Therefore, the magnitude of the force that each wire exerts on the other will increase by a factor of two.