91Ó°ÊÓ

Imagine spinning a rod around its center. The moment of inertia tells us how hard it is to get it spinning. It's like the rotational equivalent of mass in linear motion—it measures the resistance an object has to changes in its motion. For a thin rod or blade spinning around its center, the moment of inertia \( I \) can be calculated using the formula:

• \( I = \frac{1}{12} m L^2 \)

where \( m \) is the mass and \( L \) is the length of the rod. In our case, these values are \( 24.0 \, \text{kg} \) for mass and \( 2.5 \, \text{m} \) for length, giving us a moment of inertia of \( 12.5 \, \text{kg} \cdot \text{m}^2 \). This tells us how "stubborn" the blade is when it comes to starting or stopping its spin.

Torque is like a rotational force. It’s what makes things start spinning, stop spinning, or spin faster. When you apply a force offset from a pivot or axis, you create torque. In this problem, the technician applies a force of \( 35 \, \text{N} \) at a distance of \( 1 \, \text{m} \) from the axle, creating a torque calculated as:

• \( \tau = rF = 1 \, \text{m} \times 35 \, \text{N} = 35 \, \text{N} \cdot \text{m} \)

Torque is crucial in understanding how rotational motion will change. Here, the applied torque works against the frictional torque which opposes the motion.

Friction is often a party crasher in the world of motion. In terms of rotational motion, friction acts as a stumbling block to the spinning object. It produces an opposing torque that tries to slow things down. For our spinning blade, this friction creates a torque of \( 5 \, \text{N} \cdot \text{m} \). Frictional torque needs to be considered in the net torque calculations, which will affect how much the blade can accelerate. Friction always works against motion, so it essentially reduces the effectiveness of the applied force by producing counter-torque.

Rotational motion deals with objects spinning around an axis. It's a lot like linear motion, but instead of going in straight lines, it involves circular paths. Key elements include angular velocity, angular acceleration, and torque. In our exercise, we are concerned with angular acceleration \( \alpha \), which shows how quickly the rotation speed is changing. The formula used to determine this is:

• \( \alpha = \frac{\tau_\text{net}}{I} \)

where \( \tau_\text{net} \) is the net torque and \( I \) is the moment of inertia. For the rotating blade, the net torque is \( 30 \, \text{N} \cdot \text{m} \) after considering the opposing friction. Solving gives us an angular acceleration of \( 2.4 \, \text{rad/s}^2 \), determining how fast the technician can make the blade spin around its axle.