The string constrains the rotational and translational motion of the falling cylinder, given that it doesn't slip. what is the relationship between the magnitude of the angular velocity ω and that of the velocity v of the center of mass of the cylinder? express ω in terms of v and r.

First we need to find the acceleration. torque on cylinder τ = T * r where T is the string tension; T = m(g - a) where a is the acceleration of the cylinder. Then τ = m(g - a)r But also τ = Iα. For a solid cylinder, I = ½mr², and if the string doesn't slip, then α = a / r, so τ = ½mr² * a/r = ½mra. Since τ = τ, we have m(g - a)r = ½mra → m, r cancel, leaving g - a = ½a g = 3a/2 a = 2g/3 where g, of course, is gravitational acceleration. We know that v(t) = a*t, so for our cylinder v(t) = 2gt / 3 ◄ linear velocity and ω = v(t) / r = 2gt / 3r ◄ angular velocity

By dividing you can find the answer

The low lands i believe