Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)
The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)
Moment of Inertia refers to:
- the quantity expressed by the body resisting angular acceleration.
- It the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.
The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)
here We note that the,
In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.
The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:
I(edge) = I (center of mass) + md^2
d be the distance from an axis through the object’s center of mass to a new axis.
I2(edge) = 1/3 (m*L^2)
learn more about moment of Inertia here:
<u>brainly.com/question/14226368</u>
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Johann Strauss II
hope this helps
Answer:
<em>The frequency of of the note = 131 Hz.</em>
Explanation:
<em>Frequency:</em><em> Frequency can be defined as the number of complete oscillation completed by a wave in one seconds. The S.I unit of frequency is Hertz ( Hz)</em>
v = λf ............................ Equation 1
Making f the subject of the equation,
f = v/λ .......................... Equation 2
Where v = Speed, λ = wavelength, f = frequency
<em>Given: v = 343 m/s, λ = 2.62 m.</em>
<em>Substituting these values into equation 2</em>
<em>f = 343/2.62</em>
<em>f = 131 Hz</em>
<em>Thus the frequency of of the note = 131 Hz.</em>
