A spinning top is the answer
F=ma
F= 4x1.2
F= 4.8 N
F= 4gsin30 - Friction
Friction= 19.6 - 4.8 N
Friction= 14.8 N
Friction= u x 4gcos30
14.8 / 4gcos30 = u
u= 0.43596...
u= 0.44
coefficient is 0.44
Answer:
a) τ = 0.672 N m
, b) θ = 150 rad
, c) W = 100.8 J
Explanation:
a) for this part let's start by finding angular acceleration, when the angular velocity stops it is zero (w = 0)
w = w₀ + α t
α = -w₀ / t
α = 120 / 2.5
α = 48 rad / s²
The moment of inertia of a cylinder is
I = ½ M R²
Let's calculate the torque
τ = I α
τ = ½ M R² α
τ = ½ 2.8 0.1² 48
τ = 0.672 N m
b) we look for the angle by kinematics
θ = w₀ t + ½ α t2
θ = ½ α t²
θ = ½ 48 2.5²
θ = 150 rad
c) work in angular movement
W = τ θ
W = 0.672 150
W = 100.8 J
Given:
I₁ = 0.70 kg-m², the moment of inertia with arms and legs in
I₂ = 3.5 kg-m², the moment of inertia with arms and a leg out.
ω₁ = 4.8 rev/s, the angular speed with arms and legs in.
That is,
ω₁ = (4.8 rev/s)*(2π rad/rev) = 30.159 rad/s
Let ω₂ = the angular speed with arms and a leg out.
Because momentum is conserved, therefore
I₂ω₂ = I₁ω₁
ω₂ = (I₁/I₂)ω₁
= (0.7/3.5)*(30.159)
= 6.032 rad/s
ω₂ = (6.032 rad/s)*(1/(2π) rev/rad) = 0.96 rev/s
Answer: 0.96 rev/s
