(c) A small ball of mass 0.75 kg is attached to one end of a 1.25 m long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is 30° from the vertical, what is the magnitude of the gravitational torque calculated about the pivot?

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**Problem (c): Calculation of Gravitational Torque in a Pendulum**

A small ball of mass 0.75 kg is attached to one end of a 1.25 m long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is 30° from the vertical, what is the magnitude of the gravitational torque calculated about the pivot?

To calculate the gravitational torque (\(\tau\)) about the pivot, we can use the formula:

\[ \tau = r \cdot F \cdot \sin(\theta) \]

where:
- \( r \) is the length of the rod (1.25 m)
- \( F \) is the force due to gravity acting on the mass (mg), where \( m \) is the mass (0.75 kg) and \( g \) is the acceleration due to gravity (approximately 9.8 m/s²)
- \(\theta\) is the angle from the vertical (30°)

First, calculate the force \( F \):
\[ F = m \cdot g \]
\[ F = 0.75 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \]
\[ F = 7.35 \, \text{N} \]

Next, calculate the torque:
\[ \tau = 1.25 \, \text{m} \cdot 7.35 \, \text{N} \cdot \sin(30°) \]
\[ \sin(30°) = 0.5 \]
\[ \tau = 1.25 \, \text{m} \cdot 7.35 \, \text{N} \cdot 0.5 \]
\[ \tau = 4.59375 \, \text{Nm} \]

Therefore, the magnitude of the gravitational torque about the pivot is approximately **4.59 Nm**.
Transcribed Image Text:**Problem (c): Calculation of Gravitational Torque in a Pendulum** A small ball of mass 0.75 kg is attached to one end of a 1.25 m long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is 30° from the vertical, what is the magnitude of the gravitational torque calculated about the pivot? To calculate the gravitational torque (\(\tau\)) about the pivot, we can use the formula: \[ \tau = r \cdot F \cdot \sin(\theta) \] where: - \( r \) is the length of the rod (1.25 m) - \( F \) is the force due to gravity acting on the mass (mg), where \( m \) is the mass (0.75 kg) and \( g \) is the acceleration due to gravity (approximately 9.8 m/s²) - \(\theta\) is the angle from the vertical (30°) First, calculate the force \( F \): \[ F = m \cdot g \] \[ F = 0.75 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \] \[ F = 7.35 \, \text{N} \] Next, calculate the torque: \[ \tau = 1.25 \, \text{m} \cdot 7.35 \, \text{N} \cdot \sin(30°) \] \[ \sin(30°) = 0.5 \] \[ \tau = 1.25 \, \text{m} \cdot 7.35 \, \text{N} \cdot 0.5 \] \[ \tau = 4.59375 \, \text{Nm} \] Therefore, the magnitude of the gravitational torque about the pivot is approximately **4.59 Nm**.
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