Naldo6 said:A ball of mass 2.10 kg and radius 0.145 m is released from rest on a plane inclined at an angle theta=38.0 degrees with respect to the horizontal. How fast is the ball moving (in m/s) after it has rolled a distance d=1.50 m? Assume that the ball rolls without slipping, and that its moment of inertia about its center of mass is 1.40×10-2 kgm2.
The force acting on a sphere rolling down an incline is the force of gravity, which pulls the sphere downwards towards the center of the Earth. This force is proportional to the mass of the sphere and the acceleration due to gravity.
The angle of the incline affects the speed of the sphere by changing the component of the force of gravity that acts along the incline. As the angle increases, the component of gravity acting down the incline increases, causing the sphere to accelerate faster.
The height of the incline has a direct relationship with the final velocity of the sphere. As the incline height increases, the potential energy of the sphere also increases, leading to a higher final velocity as the sphere rolls down the incline.
The mass of the sphere does not affect its acceleration down the incline. The force of gravity acting on the sphere is directly proportional to its mass, so the acceleration of the sphere remains constant regardless of its mass.
The rolling resistance of the sphere on the incline can be affected by several factors, including the surface of the incline (friction), the shape and material of the sphere, and any external forces acting on the sphere (such as air resistance).