Hi I'm kinda new here, and hope someone can please help me out.
A person exerts a tangential force of 38.1 N on the rim of a disk-shaped merry-go-round of radius 2.74 m and mass 207 kg.
If the merry-go-round starts at rest, what is its angular speed after the person has rotated it through an angle of 32.5degrees?
Any help would be appreciated!
Torque= I alpha
38.1*2.74=I alpha
For I,use a solid disk.
solve for angular acceleation, alpha
wf^2 = 2*alpha*displacement, where displacement is 32.5deg in RADIANS
To find the angular speed of the merry-go-round after the person has rotated it through an angle of 32.5 degrees, you can use the principle of conservation of angular momentum.
The formula for angular momentum is given by L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
First, you need to find the moment of inertia of the merry-go-round. For a disk-shaped object, the moment of inertia is given by I = (1/2)mr^2, where m is the mass of the object and r is the radius.
Next, calculate the initial angular momentum. Since the merry-go-round starts from rest, the initial angular momentum is zero.
Once you have the moment of inertia and the initial angular momentum, you can rearrange the formula to solve for the final angular speed:
L = Iω
0 = (1/2)mr^2 ω
Solving for ω, you get:
ω = 0 / ((1/2)mr^2)
ω = 0
Therefore, the angular speed of the merry-go-round after the person has rotated it through an angle of 32.5 degrees is 0.
This means that the merry-go-round remains at rest after the person rotates it.