A solid sphere of mass m and radius r rolls down an inclined plane without slipping. Find the speed of its centre of mass when the sphere reaches the bottom
To solve this problem, we can use the principle of conservation of energy. If the sphere rolls down the inclined plane without slipping, the potential energy it loses will be converted into both kinetic energy of its center of mass and rotational kinetic energy.
Let's assume that the height of the inclined plane is h. The potential energy at the top of the plane is mgh, where g is the acceleration due to gravity. At the bottom of the plane, the potential energy is zero.
The kinetic energy of the center of mass is given by the equation: KE = (1/2)mv², where m is the mass of the sphere and v is the speed of its center of mass.
The rotational kinetic energy of the sphere is given by the equation: KE_rotational = (1/2)Iω², where I is the moment of inertia and ω is the angular velocity of the sphere.
For a solid sphere rolling without slipping, the moment of inertia is I = (2/5)mr² and the angular velocity is ω = v/r.
Therefore, the total mechanical energy of the sphere is expressed as follows:
mgh = (1/2)mv² + (1/2)(2/5)mr²(v/r)²
mgh = (1/2)mv² + (1/5)mv²
mgh = (7/10)mv²
Simplifying the above equation, we find:
v² = (10/7)gh
v = sqrt((10/7)gh)
Thus, the speed of the center of mass of the solid sphere when it reaches the bottom of the inclined plane is given by:
v = sqrt((10/7)gh).