Answer:
Step-by-step explanation:
Mass of the reel is given as,
M = 30kg
Radius of gyration about A is given as,
Ka = 120mm = 0.12m
Mass of the cylinder suspended
Mc = 40kg
Then, weight of the cylinder is
W = mg = Mc × g
W = 40 × 9.81 = 392.4 N
The exert force
P = 300N
Radius of the reel
R = 150mm = 0.15m
r = 75mm = 0.075m
Check attached for diagram of this problem
A. Moment of inertial of the reel?
We will assume the reel is a thin hoop shape, so we will use the thin hoop shape formula
I = M•Ka²
I = 30 × 0.12²
I = 0.432 kgm²
Therefore, Moment of inertia of reel is 0.432 kg.m²
B. Velocity if the cylinder after it moves 2m?
Applying torque equation
Στ = Iα
Where
α is angular acceleration
I is moment of inertia
τ is the torque..
Where τ = F × r
The two force acting on the reel is the weight of the cylinder and it is acting downward and the force applied
Then, taking torque about point A
Στ = Iα.
P × R — W × r = Iα
300 × 0.15 - 392.4 × 0.075 = 0.432 × α
45 — 29.43= 0.432α
15.57 = 0.432α
α = 15.57 / 0.432
α = 36.042 rad/s²
Now, The angle turned by the reel when the cylinder moves 2m above can be determined using
S= rθ
Then, θ = S/r
θ = 2 / 0.075
θ = 26.667 rad
Initially, the reel is at rest, then, the initial angular velocity is 0
ωo = 0 rad/s
Now, apply the kinematic equation and calculate the final angular velocity of the reel,
ω² = ωo² + 2αθ
ω² = 0² + 2 × 36.042 × 26.667
ω² = 0 + 1922.24
ω = √1922.24
ω = 43.84 rad/s
Now, using the relationship between linear velocity and angular velocity
Then, the velocity of cylinder
V = rω
V = 0.075 × 43.84
V = 3.2883 m/s
Therefore, Velocity of Cylinder is 3.288 m/s.
C. Time taken to travel 2m
From equation of circular motion
ω = ωo + αt
43.84 = 0 + 36.042t
43.84 = 36.042t
t = 43.84 / 36.042
t = 1.216 seconds
The time taken to travel 2m is 1.216 seconds