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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Answer:
Velocity = 3.25[m/s]
Explanation:
This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.
In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.
The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"
h2 is located at point 2 and it will be zero.
![(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]](https://tex.z-dn.net/?f=%28P_%7B1%7D%20%2B%5Cfrac%7Bv_%7B1%7D%5E%7B2%7D%20%7D%7B2g%7D%20%2Bh_%7B1%7D%20%29%3D%28P_%7B2%7D%20%2B%5Cfrac%7Bv_%7B2%7D%5E%7B2%7D%20%7D%7B2g%7D%20%2Bh_%7B2%7D%20%29%5C%5CP_%7B1%7D%20%3DP_%7B2%7D%20%5C%5Cv_%7B1%7D%3D0%5C%5Ch_%7B2%7D%20%3D0%5C%5Cv_%7B2%7D%3D%5Csqrt%7B0.54%2A9.81%2A2%7D%5C%5Cv_%7B2%7D%3D3.25%5Bm%2Fs%5D)
Its b i literally have had this exact question
Answer:
The required angle is (90-25)° = 65°
Explanation:
The given motion is an example of projectile motion.
Let 'v' be the initial velocity and '∅' be the angle of projection.
Let 't' be the time taken for complete motion.
Let 'g' be the acceleration due to gravity
Taking components of velocity in horizontal(x) and vertical(y) direction.
= v cos(∅)
= v sin(∅)
We know that for a projectile motion,
t =
Since there is no force acting on the golf ball in horizonal direction.
Total distance(d) covered in horizontal direction is -
d =
×t = vcos(∅)×
=
.
If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -
α +β = 90° as-
d =
=
=
=
.
∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.
∴ If α = 25° , then
β = 90-25 = 65°
∴ The required angle is 65°.
For an object moving in a path that's a circle or a part of one,
the centripetal force acts in the direction toward the center of
the circle. That direction is perpendicular to the way the object
is moving.