Question

A solid cylinder of mass m and radius r rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be:

a) √(2gh/r)
b) √(gh/2r)
c) √(gr/2h)
d) √(2gr/h)

Answer 1

The angular velocity of a solid cylinder reaching the bottom of an inclined plane is a. √(2gh/r), obtained by applying conservation of energy principles to the cylinder's translational and rotational kinetic energies.

Step-by-step explanation:

The question is asking for the angular velocity of a solid cylinder when it reaches the bottom of an inclined plane. The conservation of energy principle is used to find the angular velocity of an object that rolls without slipping. It states that when a solid cylinder rolls down an incline from a height h, its potential energy is converted into both translational and rotational kinetic energy.

For a cylinder rolling without slipping, we can write the total kinetic energy as KEtotal = (1/2)mv2 + (1/2)Iω2, where I is the moment of inertia, v is the linear velocity, and ω is the angular velocity. Since the moment of inertia for a solid cylinder is (1/2)mr2, the conservation of energy gives us mgh = (1/2)mv2 + (1/4)mr2ω2. Because v = ωr, ω can be isolated to be ω = √(4gh/3r), which simplifies to ω = √(2gh/r). Therefore, the correct answer is Option (a) √(2gh/r).