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Question 1198759: A products lift accelerates from rest to 0.16 m/s2
in 32 seconds. It then moves at constant velocity
for 33 seconds and then decelerates to rest in 15 seconds as . Lift pulley is 30cm
diameter carrying load 30 kg.
Apply dimensional analysis techniques and solve:
a) The linear velocity of the pulley.
b) The angular velocity of the cable.
c) The angular acceleration of the pulley.
d) The linear acceleration of cable.
e) The torque applied to the shaft.
State the formulae used to show that all values/units used are homogeneous. Given that :
v = vo + at , where , v ,is velocity , vo , is the initial velocity , a , is the acceleration and t is time.
Apply dimensional analysis techniques to develop two equations for power in terms of linear
velocity and angular velocity, from the formulae used above.
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **a) Linear Velocity of the Pulley**
* **Acceleration Phase:**
* v = u + at
* v = 0 + (0.16 m/s²) * (32 s)
* v = 5.12 m/s
* **Constant Velocity Phase:**
* The linear velocity remains constant at 5.12 m/s during this phase.
* **Deceleration Phase:**
* v = u + at
* 0 = 5.12 m/s + a * (15 s)
* a = -5.12 m/s / 15 s
* a = -0.3413 m/s²
* **Linear Velocity of the Pulley:** 5.12 m/s
**b) Angular Velocity of the Cable**
* **Linear Velocity (v):** 5.12 m/s
* **Radius of the Pulley (r):** 0.30 m / 2 = 0.15 m
* **Angular Velocity (ω):** ω = v / r
* ω = 5.12 m/s / 0.15 m
* ω = 34.13 rad/s
**c) Angular Acceleration of the Pulley**
* **Linear Acceleration (a):** 0.16 m/s²
* **Radius of the Pulley (r):** 0.15 m
* **Angular Acceleration (α):** α = a / r
* α = 0.16 m/s² / 0.15 m
* α = 1.067 rad/s²
**d) Linear Acceleration of the Cable**
* **During Acceleration:** 0.16 m/s²
* **During Constant Velocity:** 0 m/s²
* **During Deceleration:** -0.3413 m/s²
**e) Torque Applied to the Shaft**
* **Torque (τ):** τ = I * α
* where I is the moment of inertia of the pulley and α is the angular acceleration.
* **Assuming the pulley is a solid disk:**
* I = (1/2) * M * r²
* where M is the mass of the pulley and r is the radius.
* **To find the mass of the pulley, we need the density of the material.**
**Dimensional Analysis**
* **v = u + at**
* [L/T] = [L/T] + [L/T²] * [T]
* [L/T] = [L/T]
* **ω = v / r**
* [rad/s] = [L/T] / [L]
* [1/s] = [1/s]
* **α = a / r**
* [rad/s²] = [L/T²] / [L]
* [1/s²] = [1/s²]
* **τ = I * α**
* [N.m] = [kg.m²] * [rad/s²]
* [kg.m/s²] * [m] = [kg.m²/s²] * [m]
* [kg.m²/s²] = [kg.m²/s²]
**Power Equations**
* **Power (P) = Force (F) * Velocity (v)**
* P = m * a * v
* [W] = [kg] * [L/T²] * [L/T]
* [kg.m²/s³] = [kg.m²/s³]
* **Power (P) = Torque (τ) * Angular Velocity (ω)**
* P = I * α * ω
* [W] = [kg.m²] * [rad/s²] * [rad/s]
* [kg.m²/s³] = [kg.m²/s³]
**Note:**
* This analysis assumes that the pulley is a solid disk.
* The mass of the pulley and the load it carries will affect the torque required and the overall power consumption.
* This analysis does not consider factors such as friction and efficiency losses.
**I hope this comprehensive analysis helps!**
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