Final answer:
To calculate the linear speed of the disk at the bottom of the incline, energy conservation principles are used, considering both translational and rotational kinetic energies. The potential energy at the top due to gravity is equated to the kinetic energies at the bottom, and solving for velocity v involves the mass, the height of the incline, and the radius of the disk.
Step-by-step explanation:
The question involves calculating the linear speed of the center of a uniform solid disk rolling down an incline, which is a problem involving the principles of rotational motion and energy conservation in physics. To solve this problem, we will use the conservation of energy, accounting for both the translational kinetic energy and the rotational kinetic energy of the disk.
The total mechanical energy at the top of the incline (potential energy) will be equal to the sum of translational and rotational kinetic energies at the bottom of the incline. The potential energy (PE) at the top is given by PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height of the incline. The height can be found using trigonometry (h = L*sin(26 degrees), where L is the length of the incline).
The kinetic energy at the bottom will have two parts: translational (½mv²) and rotational (½Iω²), where I is the moment of inertia of the disk and ω (omega) is the angular velocity. For a solid disk, the moment of inertia I is ½mR². Since the disk rolls without slipping, ω can be related to v through the relation ω = v/R, where R is the radius of the disk.
Equating the potential energy at the top to the sum of translational and rotational kinetic energies at the bottom gives us: mgh = ½mv² + ½(½mR²)(v/R)². Solving for v would yield the linear velocity of the disk's center at the bottom of the incline. Remember that the diameter was given, so the radius R is half the diameter.