Engineering, 21.04.2020 19:20 Shathep

The reel has a mass of 30 kg and a radius of gyration about A of kA = 120 mm. The suspended cylinder has a mass of 40 kg. Starting from rest, the motor M exerts a constant force of P = 300 N on the cable.  Determine the mass moment of inertia of the reel.  Determine the velocity of the cylinder after it has traveled upward 2 m.  Determine the time that was taken to travel the 2 m distance.

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Answer from: loudenalexisp56lp0

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

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The reel has a mass of 30 kg and a radius of gyration about A of kA = 120 mm. The suspended cylinder...

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