Answer:
a) α = 0.338 rad / s² b) θ = 21.9 rev
Explanation:
a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque
τ = I α
fr r = I α
Now we write the translational Newton equation in the radial direction
N- F = 0
N = F
The friction force equation is
fr = μ N
fr = μ F
The moment of inertia of a saying is
I = ½ m r²
Let's replace in the torque equation
(μ F) r = (½ m r²) α
α = 2 μ F / (m r)
α = 2 0.2 24 / (86 0.33)
α = 0.338 rad / s²
b) let's use the relationship of rotational kinematics
w² = w₀² - 2 α θ
0 = w₀² - 2 α θ
θ = w₀² / 2 α
Let's reduce the angular velocity
w₀ = 92 rpm (2π rad / 1 rev) (1 min / 60s) = 9.634 rad / s
θ = 9.634 2 / (2 0.338)
θ = 137.3 rad
Let's reduce radians to revolutions
θ = 137.3 rad (1 rev / 2π rad)
θ = 21.9 rev
Answer: -4.4 m/s
Explanation:
This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum 
must be equal to the final momentum 
:
(1)
Where:
(2)
(3)
is the mass of the child
is the initial velocity of the child
is the mass of the adult
is the initial velocity of the adult (it is sitting still)
is the final velocity of the child
is the final velocity of the adult
Substituting (2) and (3) in (1):
(4)
Isolating :
(5)
(6)
Finally:
This means the velocity of the child is in the opposite direction
