If the pipe rolls without slipping, what will be its speed at the base of the in
ID: 1293353 • Letter: I
Question
If the pipe rolls without slipping, what will be its speed at the base of the incline?
What will be its total kinetic energy at the base of the incline?
What minimum value must the coefficient of static friction have if the pipe is not to slip?
A thin, hollow 0.540-kg m long. If the pipe rolls without slipping, what will be its speed at the base of the incline? What will be its total kinetic energy at the base of the incline? What minimum value must the coefficient of static friction have if the pipe is not to slip? incline 5.90m cm starts rolling (from rest) down a 16.5? kg section of pipe of radius 10.0 cmExplanation / Answer
The first part involves conservation of energy. Since there is no slipping, the energy from the work done by friction gives the pipe angular KE.
For a thin hollow cylinder, I=m*r^2
When the cylinder rolls without slipping, ?=v/r
The loss in PE m*g*5.9*sin16.5
becomes translational and angular KE of the pipe
.5*m*v^2+.5*I*?^2
plug in I=m*r^2 and ?=v/r
.5*m*v^2+.5*m*r^2*v^2/r^2
simplify
.5*m*v^2*2
build the energy equation
m*g*5.9*sin16.5=.5*m*v^2*2
solve for v
v=sqrt(g*5.9*sin16.5)
v=4.05 m/s
The total KE is
m*g*5.9*sin16.5
8.97 J
For