91Ó°ÊÓ

Torque is a pivotal concept when dealing with rotational motion. Think of it as the rotational equivalent of force. It's what makes an object spin. In simpler terms, torque is what causes an object to start, stop, or change its rotational motion around an axis. The formula for torque \( \tau \) is:

• \( \tau = r \cdot F \cdot \sin(\theta) \)
• where \( r \) is the distance from the pivot point to the point where the force is applied.
• \( F \) is the force applied.
• \( \theta \) is the angle between the force vector and the lever arm.

In our exercise, the gravitational force acts on the center of the metre stick, producing torque around the hinge. This torque fosters angular acceleration once the string is cut, setting the stick in motion.

The moment of inertia is to rotational motion what mass is to linear motion. It is a measure of an object's resistance to changes in its rotation. The larger the moment of inertia, the harder it is to change its rotational state. For a uniform rod with a pivot at one end, like the metre stick in our exercise, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{3} M L^2 \] This equation factors in:

• \( M \), the mass of the object, and
• \( L \), the length of the object.

A higher moment of inertia implies more effort needed to start or stop rotation. When the string is cut, the moment of inertia affects how quickly the stick starts to rotate.

Angular acceleration is how quickly an object speeds up or slows down its rotation. It's similar to linear acceleration but for rotational movements. In the context of the exercise, it's what we calculate to find how the stick begins its motion when the string is cut.
Angular acceleration \( \alpha \) is directly related to the applied torque \( \tau \) and the object's moment of inertia \( I \):

• \( \tau = I \alpha \)

Rearranging the formula gives us how to calculate angular acceleration:
\[ \alpha = \frac{\tau}{I} \] By substituting in the values for torque and moment of inertia, we work out the stick's initial angular acceleration. For our metre stick, this yields \( \alpha = \frac{3g}{2} \), helping us conclude the exercise correctly. Angular acceleration is crucial as it determines how rapidly an object ramps up its spin from rest.