Answer:
onservation of energy
U top = K bottom
(m + m)*g*L = 1/2*I*?^2 where I = m*(L/2)^2 + m*L^2 = 1.25*m*L^2
So 2m*g*L = 1/2*1.25*m*L^2*?^2
So ? = sqrt(3*g*/(1.25*L) ) = sqrt(12g/5L)
Center.........................
Answer:
Explanation:
As in any sample you will have 75.8% of Cl-35 iosotopes and 24.3% of Cl-37 iosotopes you can get the average atomic mass as:
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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