A solid cylinder?

Question

with a mass of 3.49 kg and a radius of 0.075 m starts from rest at a height of 3.60 m and rolls down a 43.8 degree slope. Acceleration of gravity is 9.81 m/s^2.

What is the translational speed of the cylinder when it leaves the incline in units of m/s?

Answer 1

The angle of the slope is irrelevant, and we need to assume that the cylinder is rolling without slippage. In that case, you can do a simple energy balance. If PE is gravitational potential energy, KT is translational kinetic energy, and KR is rotational kinetic energy, then PE (initial) = KT + KR (final). PE is easily calculated as mgh, where you were given mass m and height h, as well as g, the acceleration due to gravity. If the cylinder has a tranlational speed v at the bottom of the slope, the translational kinetic energy is 0.5*m*v^2. Rotational kinetic energy is a little trickier, because it's 0.5*I*w^2, where I is the moment of inertia and w is the angular velocity. Given the radius r of the cylinder, as you were, w = v/r. That's where the "no slippage" assumption comes in, because it allows you to assume that the edge of the cylinder moves at a speed equal to the translational speed. With slippage, it would move slower. I is also formulaic, I = 0.5*m*r^2. Your final formula is mgh = 0.5*m*v^2 + 0.5*(0.5*m*r^2)*(v/r)^2. Note that you can simplify the KR term to eliminate r, and then combine it with the KT term before solving for v, your only unknown.