A merry-go-round rotates at the rate of 0.2 rev/s with an 88 kg man standing at a
point 2.1 m from the axis of rotation.
What is the new angular speed when the
man walks to a point 0 m from the center?
Consider the merry-go-round is a solid 75 kg
cylinder of radius of 2.1 m.
Answer in units of rad/s.

So, the new angular speed when the man walks to a point 0 m from the center (in rad/s) is approximately 0.87π rad/s.

Hi ! I will help you with this problem. This problem will mostly adopt the principle of rotational dynamics, especially angular momentum. When there is a change in the mass or radius of rotation of an object, it will not affect its angular momentum, only its angular velocity will change. Therefore, the law of conservation of angular momentum can be written as :

With the following condition :

Previously, we assumed that when rotating, only merry-go-round that experienced inertia so the man will only add to the mass of the merry-go-round. Perhaps, having a man at the end of it will change the slope (direction of torque), but not change the value of angular momentum. So, when this person is in the middle or on the central axis (read: 0 m from the center), the man no longer had any effect on the weight gain.

We know that :

What was asked :

Step by step :

So, the new angular speed when the man walks to a point 0 m from the center (in rad/s) is approximately 0.87π rad/s

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000