Angular velocity = (75x2pie)/60
=2.5pie ras^-1
linear velocity(or speed) at end of string, v = radius x angular velocity
v= 0.5 x 2.5pie
v=3.93 ms^-1
tension of string (I beleve is centeral force aplied by string), F= (mv^2)/r
F= (0.2 x 3.93^2)/0.5
F=6.18 N
(sorry if wrong)
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
The gravitational force will double as well because
F=mg
