Question

A uniform thin-walled cylinder and a uniform solid cylinder are rolling horizontally without slipping. The speed of the thin-walled cylinder is v. The cylinders encounter an incline that they climb without slipping. If the maximum height they reach is the same, find the initial speed v of the solid cylinder.

Answer 1

The initial speed v of the solid cylinder is found using conservation of energy and the moment of inertia for a solid cylinder, resulting in v = √(4/3gh), where h is the maximum height reached on the incline.

Step-by-step explanation:

To find the initial speed v of the solid cylinder, we can use the conservation of energy principle. Since both cylinders reach the same height, they both convert their initial kinetic energy into the same amount of potential energy. For the solid cylinder, the initial kinetic energy is the sum of its translational and rotational kinetic energy.

The translational kinetic energy is given by (1/2)mv2, and the rotational kinetic energy by (1/2)Iω2, where I is the moment of inertia and ω (omega) is the angular velocity. For a solid cylinder, I = (1/2)mR2 and ω = v/R, so the rotational kinetic energy is (1/4)mv2. Thus, the total initial kinetic energy of the solid cylinder is (1/2)mv2 + (1/4)mv2 = (3/4)mv2.

The potential energy at the maximum height h is mgh. Setting the initial kinetic energy equal to the potential energy, we get (3/4)mv2 = mgh. Solving for v, we find v = √(4/3gh). Since the height h is the same for both the solid and thin-walled cylinders, and assuming g (acceleration due to gravity) is constant, we can conclude that the initial speed v of the solid cylinder is √(4/3gh), where h is the maximum height reached.