91Ó°ÊÓ

Newton's Second Law is a fundamental principle in physics that explains how the velocity of an object changes when it's subjected to an external force. It is expressed by the equation \( F = ma \), where \( F \) is the net force acting on the object, \( m \) is the mass of the object, and \( a \) is the linear acceleration it undergoes.

In our problem scenario, the horizontal force applied to the cylinder (\( F_{\text{app}} = 12 \text{ N} \)) causes it to accelerate. The cylinder's resistance to change in motion, due to its mass \( m \) of 10 kg, combined with the frictional force, which ensures it rolls smoothly, allows us to apply Newton's Second Law. We have:- Applied force \( F_{\text{app}} \)- Frictional force \( f \)

The net force equation is formed as \( F_{\text{app}} - f = ma \). Solving this gives the linear acceleration \( a \) for the cylinder's center of mass.

In rotational motion, torque is the rotational equivalent of force. Torque produces angular acceleration in a body. The amount of torque \( \tau \) depends on the force applied and the distance from the pivot point, calculated by \( \tau = fR \), where \( R \) is the radius of the rotation path.

For our cylinder, the frictional force \( f \) creates a torque about its center of mass. The moment of inertia (\( I \)) for a solid cylinder, which represents the mass distribution relative to the axis of rotation, is given by \( I = \frac{1}{2}mR^2 \). This formula accounts for how difficult it is to change the cylinder's rotational state.

• Torque \( \tau = fR \)
• Moment of Inertia \( I = \frac{1}{2}mR^2 \)

Using these relations, the angular acceleration \( \alpha \) can be found using \( \tau = I \alpha \). This demonstrates how torque and moment of inertia are vital to understanding rotational motions.

Linear and angular accelerations are closely related when dealing with rolling objects. Linear acceleration \( a \) measures how quickly the velocity of the cylinder's center of mass changes, while angular acceleration \( \alpha \) concerns how fast the cylinder's rotation speeds up.

For a rolling cylinder, without slipping, these accelerations are connected through the relationship \( a = \alpha R \), where \( R \) is the cylinder's radius. Using this bridge between linear and angular motion allows us to solve problems involving objects that roll instead of sliding:

• Linear acceleration (\( a \)) affects the motion along a path.
• Angular acceleration (\( \alpha \)) affects how quickly an object rotates.

By applying this relationship in our problem, and knowing the radius and the calculated linear acceleration, we determined the angular acceleration. This reflects how the cylinder's movement on the surface is governed by these interconnected accelerations.