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>
#SPJ4
It goes in the downward direction
Answer:
Explanation:
Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.
At any distance x from point A mass density
Lets take element mass at distance x
dm =λ dx
mass moment of inertia
So total moment of inertia
By putting the values
By integrating above we can find that
Now to find location of center mass
Now by integrating the above
So mass moment of inertia and location of center of mass
