Recent content by Jeremymu1195

Cool, thank you!

Yes, I am interested in how the whole setup reacts to a torque. So the moment of inertia of the whole assembly would be the sum of the moments of the individual parts?

What is the overall moment of inertia of many cylinders in contact, where each is rotated about an axis through its center of mass. For example, a set of rollers. (See picture)

Homework Statement A mass m with velocity v0 makes an inelastic collision with second mass M that is suspended by a string of length L. The velocity v0 is perpendicular to the vertical string. After the collision the combined masses, m+M, rotate in a vertical plane around the point of...

∑Fx=ma (Newton's Second Law) F(Push)=ma The only force acting in the direction of motion is the push. We can also write Newton's second law for the y-direction, but it doesn't tell us anything. ∑Fy=N-mg=0 →N=mg. Back to the x-direction, we found that a=F(Push)/m by solving Newton's...

My attempt, N2 for the y-direction reads, Ny-mg=ma_y or, Ny-mg=-mA(2π/λ)^2 v^2 cos(2πx/λ). When the wheel loses contact, Ny=0, -mg=-mA(2π/λ)^2 v^2 cos(2πx/λ) so, v^2=g/(A(2π/λ)^2 cos(2πx/λ)) (*) but when x=nλ, cos(2πx/λ)=1 and a this corresponds to local maxima on the cosine...

The Normal force is 0 when the wheel loses contact with the surface

In the x-direction: ∑F=Fpull-Nx=0 In the y-direction: ∑F=Ny-mg=ma where a=(du/dx)(dx/dt)=-A(2π/λ)^2 v^2 cos(2πx/λ).

By statement of the problem, the velocity in the x-direction is constant, call it v. Since y=A cos(2πx/λ), The velocity in the y-direction (call it u) is given by u=dy/dt. By the chain rule, v=(dy/dx)(dx/dt)= -A(2π/λ)sin(2πx/λ)(dx/dt). But dx/dt = v. So, u = -A(2π/λ)v sin(2πx/λ). The sum of...

Homework Statement A wheel is pulled over a frictionless washboard surface with a constant horizontal velocity component. The equation of the surface is y=Acos(2πx/λ) where A is the height of the surface and x is the horizontal distance along the surface. At what value v_x does the wheel begin...