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Torque is a very important concept in angular dynamics because it helps us understand how forces cause objects to rotate. Think of it as the rotational equivalent of a force in linear motion. Torque, represented by the symbol \( \tau \), is defined by the formula:

• \( \tau = r \times F \)

Here, \( r \) is the distance from the pivot point to where the force is applied (the lever arm), and \( F \) is the force applied perpendicular to the lever arm. In the exercise, the stick is subject to a gravitational force acting at its center of mass, and the torque about pivot point B can be calculated as\[ \tau = \frac{l}{2} \times Mg = \frac{Mgl}{2} \]Where \( \frac{l}{2} \) is the distance from point B to the stick's center of mass. Hence, this torque causes the stick to start rotating as soon as it's released.

Once we know the torque, the next step is to calculate angular acceleration, often denoted as \( \alpha \). Angular acceleration is the rate at which the angular velocity of a rotating object changes with time. According to Newton's Second Law for rotation, we have:

• \( \tau = I \alpha \)

Solving for \( \alpha \), we find\[ \alpha = \frac{\tau}{I} \]Where \( I \) is the moment of inertia of the stick. For our stick, which rotates about one end, the moment of inertia \( I \) is \( \frac{1}{3}Ml^2 \). Substituting the values, we can calculate the angular acceleration as:\[ \alpha = \frac{\frac{Mgl}{2}}{\frac{1}{3}Ml^2} = \frac{3g}{2l} \]This angular acceleration is what causes the stick to change how quickly it rotates as it starts falling.

The moment of inertia \( I \) is a property of an object that measures its resistance to changes in its rotation. It plays a similar role to mass in linear motion. The moment of inertia depends on the mass of the object and how that mass is distributed relative to the axis of rotation.For this exercise, we consider the stick as a rod rotating about one of its ends, where the moment of inertia is described by:\[ I = \frac{1}{3}Ml^2 \]The larger the moment of inertia, the harder it is to start or stop the rotation. This is why the mass distribution is crucial: even if two objects have the same mass, they can have different moments of inertia depending on how the mass is spread out around the axis of rotation.Understanding this concept clarifies why the angular acceleration in the previous section is defined as \( \alpha = \frac{\tau}{I} \). It shows the relationship between the distribution of mass and the rotational motion.

The center of mass is a key point in understanding how an object's mass is distributed. It acts as if all the mass of the object were concentrated at that point. For a uniform rod like our stick, the center of mass is at its midpoint.In the exercise, the center of mass is crucial when calculating the torque and the vertical acceleration. Upon release, the center of mass accelerates as if it were a particle with mass \( M \) under the influence of gravity. But due to angular motion, the linear acceleration at the center of mass is derived from the angular acceleration:\[ a_{cm} = \alpha \cdot \frac{l}{2} = \frac{3g}{4} \]This concept helps in understanding how forces and rotations affect various points in a rigid body like a stick.

Newton's Second Law for Rotation extends the well-known linear second law \( F=ma \) to rotational dynamics. Its formula is

• \( \tau = I \alpha \)

It relates the net torque acting on a system to its moment of inertia and its angular acceleration. In this exercise, it's used to connect the torque due to gravitational force to the angular acceleration of the stick.From our calculated results,\[ \alpha = \frac{\tau}{I} = \frac{3g}{2l} \]One can see how crucial this law is in predicting the object's rotational dynamics. It's used to find the upward force at point B as well. The net vertical force must provide for the acceleration of the center of mass and counteract gravity, hence\[ F_{net} = M a_{cm} \]Solving leads to the force at B:\[ N = Mg - M \cdot \frac{3g}{4} = \frac{Mg}{4} \]Newton's second law creates a clearer picture of how forces work to create motion, both linearly and angularly.