The spool of cable has a mass of m and a radius of gyration of ke. The tension in the cable T is a function of time t, when t is in seconds. Neglect the mass of the unwound cable, and assume it is always at a constant radius. a.) What is the moment of Inertia of the spool? b.) What is the angular acceleration of the wheel at t (Use + for the ccw direction)? c.) How quickly is the cable being pulled at time t (Use + for to the right) in m/s? do h d₁ 0 = 60° Variable do d h kc Ethe m T Value 1m 0.5 m 2.25 m 0.77 m 100 kg 30t³ +77t N 1.81 s 135i + 0) N

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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**Physics of Spool Mechanics**

The spool of cable has a mass of \( m \) and a radius of gyration of \( k_c \). The tension in the cable \( T \) is a function of time \( t \), where \( t \) is in seconds. Neglect the mass of the unwound cable, and assume it is always at a constant radius.

**Diagram Description:**
The diagram illustrates a spool of cable resting on a horizontal surface. The spool has an outer diameter \( d_o \) and an inner diameter \( d_i \). A force due to tension \( T \) acts horizontally to the right on the cable wound around the spool.

A coordinate system is implied with horizontal and vertical directions. The distance from the ground to the center of the spool is denoted as \( h \), and there is an angle \( \theta \) of 60° between the tension in the cable and the line perpendicular to the ground.

**Parameters:**
- Outer Diameter \( d_o = 1 \, \text{m} \)
- Inner Diameter \( d_i = 0.5 \, \text{m} \)
- Height \( h = 2.25 \, \text{m} \)
- Radius of Gyration \( k_c = 0.77 \, \text{m} \)
- Mass \( m = 100 \, \text{kg} \)
- Tension in the cable \( T = 30t^3 + 77t \, \text{N} \)
- Time \( t = 1.81 \, \text{s} \)
- Force \( \vec{F} = 135t + 0j \, \text{N} \)

**Exercises:**

a) **Moment of Inertia of the Spool:**
   To find the moment of inertia \( I \) of the spool, we use the formula:
   \[
   I = m k_c^2
   \]

b) **Angular Acceleration:**
   The angular acceleration \( \alpha \) of the wheel at time \( t \) can be computed using:
   \[
   \alpha = \frac{\tau}{I}
   \]
   where \( \tau \) is the torque.

c) **Cable Pulling Speed:**
   To determine how quickly the cable is being pulled at time \( t \),
Transcribed Image Text:**Physics of Spool Mechanics** The spool of cable has a mass of \( m \) and a radius of gyration of \( k_c \). The tension in the cable \( T \) is a function of time \( t \), where \( t \) is in seconds. Neglect the mass of the unwound cable, and assume it is always at a constant radius. **Diagram Description:** The diagram illustrates a spool of cable resting on a horizontal surface. The spool has an outer diameter \( d_o \) and an inner diameter \( d_i \). A force due to tension \( T \) acts horizontally to the right on the cable wound around the spool. A coordinate system is implied with horizontal and vertical directions. The distance from the ground to the center of the spool is denoted as \( h \), and there is an angle \( \theta \) of 60° between the tension in the cable and the line perpendicular to the ground. **Parameters:** - Outer Diameter \( d_o = 1 \, \text{m} \) - Inner Diameter \( d_i = 0.5 \, \text{m} \) - Height \( h = 2.25 \, \text{m} \) - Radius of Gyration \( k_c = 0.77 \, \text{m} \) - Mass \( m = 100 \, \text{kg} \) - Tension in the cable \( T = 30t^3 + 77t \, \text{N} \) - Time \( t = 1.81 \, \text{s} \) - Force \( \vec{F} = 135t + 0j \, \text{N} \) **Exercises:** a) **Moment of Inertia of the Spool:** To find the moment of inertia \( I \) of the spool, we use the formula: \[ I = m k_c^2 \] b) **Angular Acceleration:** The angular acceleration \( \alpha \) of the wheel at time \( t \) can be computed using: \[ \alpha = \frac{\tau}{I} \] where \( \tau \) is the torque. c) **Cable Pulling Speed:** To determine how quickly the cable is being pulled at time \( t \),
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