A person exerts a tangential force of 34.8 N on the rim of a disk-shaped merry-go-round of radius 2.60 m and mass 152 kg. If the merry-go-round starts at rest, what is its angular speed after the person has rotated it through an angle of 30.4°?
(in rad/s)
To find the angular speed of the merry-go-round, we can use the conservation of angular momentum. The equation for angular momentum is:
L = Iω
where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
First, we need to find the moment of inertia of the merry-go-round. The moment of inertia for a disk-shaped object rotating about its axis is given by:
I = 0.5MR^2
where M is the mass of the merry-go-round, and R is the radius.
Plugging in the given values:
M = 152 kg
R = 2.60 m
I = 0.5 * 152 kg * (2.60 m)^2
Next, we can calculate the change in angular momentum (∆L). It is given by:
∆L = F * ∆t
where F is the tangential force and ∆t is the time interval.
Plugging in the given value:
F = 34.8 N
To calculate ∆t, we need to use the equation relating angle θ, time ∆t, and angular speed ω:
θ = ω * ∆t
Rearranging the equation, we get:
∆t = θ / ω
Plugging in the given value:
θ = 30.4° = 0.532 rad
∆t = 0.532 rad / ω
Now, let's substitute the values into the equation for ∆L:
∆L = F * (θ / ω)
We know that angular momentum is conserved, so:
∆L = I * ω - I * 0
Simplifying the equation, we have:
F * (θ / ω) = I * ω
Now, let's solve for ω:
F * θ = I * ω^2
ω = √((F * θ) / I)
Plugging in the values:
ω = √((34.8 N * 0.532 rad) / (0.5 * 152 kg * (2.60 m)^2))
Calculating this expression, we get the angular speed (ω) in rad/s.