91Ó°ÊÓ

Rolling motion is a fascinating aspect of physics that involves an object moving smoothly over a surface. For rolling without slipping, a crucial condition is that the point on the object in contact with the surface must have instantaneously zero velocity relative to the surface.
For rolling spheres, like the one in our problem, the movement is a combination of translational motion of its center of mass and rotational motion around its center of mass. This dual motion is significant since it allows the sphere to move efficiently across different inclines.

• Translational motion: The linear motion of the sphere’s center of mass.
• Rotational motion: The spinning of the sphere around its own axis, depicted by angular velocity \( \omega \).

These two aspects are related through the rolling motion condition: \( v = r\omega \), where \( v \) is the linear velocity of the center of mass and \( r \) is the radius of the sphere. The moment of inertia, \( I = \frac{2}{5}mr^2 \), plays a critical role as it influences how the mass of the sphere is distributed around its axis of rotation.

The principle of conservation of energy is a cornerstone in physics, dictating that the total energy in an isolated system remains constant. When applying this to our sphere, which rolls down the incline, we consider different forms of energy.

At the top of the incline, the sphere has gravitational potential energy, given by \( mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height. As the sphere descends, this potential energy is converted into kinetic energy. The kinetic energy is split into two parts:

• Translational kinetic energy: \( \frac{1}{2}mv^2 \)
• Rotational kinetic energy: \( \frac{1}{2}I\omega^2 \)

Using conservation of energy, we set up the equation:
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]
Substituting \( I = \frac{2}{5}mr^2 \) and \( \omega = \frac{v}{r} \), this simplifies to \( v = \sqrt{\frac{10gh}{7}} \). Thus, the final speed at the bottom depends solely on the vertical height, not the angle of the slope.

Kinematics is the branch of physics that deals with the motion of objects, without considering the forces that cause this motion. It provides tools to calculate various parameters of motion, such as time, distance, velocity, and acceleration. In our scenario, we are interested in how long it takes the sphere to roll down the incline.
To find the time taken, we consider the acceleration along the incline. The component of the gravitational force that causes acceleration is \( mg \sin(\theta) \), leading to an acceleration \( a = g \sin(\theta) \). This affects how quickly the sphere reaches the bottom.
Utilizing the kinematic equation:
\[ s = ut + \frac{1}{2}at^2 \]
Where \( u \) is the initial velocity (0 in this case), and \( s \) is the distance along the incline. Solving for \( t \), we get:
\[ t = \sqrt{\frac{2s}{g \sin(\theta)}} \]
This formula shows that a shallower plane will have a larger distance \( s \) and a smaller acceleration \( a \), resulting in a longer descent time compared to a steeper slope.