Question: 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 kinetic energy when it leaves the incline equals the potential energy it had at the top. The potential energy is M*g, where M = mass of the cylinder. The kinetic energy is .5*M*v^2+.5*I*w^2, where v=translational velocity and w = angular velocity, I = moment of inertia. w = v/r, so the formula becomes KE = .5*v^2*(M+I*v^2/r^2). Equating the energies get

M*g*h = .5*v^2(M + I/r^2); v=√[2*M*g*h/(M + I/r^2)]

The moment of inertia is given by ∫∫∫r^2dm, for a cylinder of length L, dm = rho*r*dr*dø, so the moment is ∫∫L*rho*r^3*dr*dø. (Rho is the density of the cylinder.) The angle ø is integrated from 0 to 2π leaving I = 2*π*L*rho*∫r^3*dr, which is 2*π*L*rho*(r^4)/4. However π*r^2*L*rho = M, so

I =.5*M*r^2

Putting this into the energy balance equation gives

v=√[2*M*g*h/(M+.5*M*r^2/r^2)]

v=√[2g*h/1.5)] = √[(4*g*h/3)]