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:
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The mass would be 51 kg. So b is the right answer.
The upper back muscles being worked while using a rowing machine .your upper trapezius and rhomboids located between your shoulder blades, and latissimus dorsi located beneath the armpits
Answer:
112.5 N
Explanation:
50 = GMm/r^2
Let F be the new force of attraction
F/50 = ( G(3M)(3m)/(2r)^2 ) / (GMm/r^2)
[Elimiating G,M,m,r]
F = 112.5 N