Question

A hollow cylinder (hoop) is rolling on a horizontal surface at speed v = 3.3 m/s when it reaches a 16 ? incline.

How far up the incline will it go?

How long will it be on the incline before it arrives back at the bottom?

Answer 1

Moment of inertia for hollow cylinder (hoop) is

             I = mr² ............. (1)

at base of incline:

total energy = translational kinetic energy + rotational kinetic energy

                     = (1/2)(mv²) + (1/2)(I?²)   ............. (2)

the relation between angular speed and linear speed is

                  v = r?

   angular speed ? = v/r ............ (3)

substitute the eq (3) and eqn (1) in eqn (2), we get

       energy = (1/2)(mv²) + (1/2)((mr²)(v/r)²)

                   = (1/2)(mv²) + (1/2)(mv²)

                   = mv²   ................... (4)

the total mechanical energy at the bottom of (h = 0 m) is the same as the

total mechanical energy at the top(h = h₀) .

          i.e., mv² = mgh₀

                     v² = h₀

                     h₀ = (v² /g)   ................. (5)

from figure,

           sin16° = h₀/x

           h₀ = x sin16° .............. (6)

compare eq (5) and (6), we get

         x sin16° = (v² /g)

         x = (v² /g*sin16° ) ............. (7)

substitute the given data in eqn(7), we get

        x = 4.027 m

        time t (to reach the top) = (total distance) /(average velocity)

= x/(v/2) = (2)(4.027)/(3.3) = 2.440 s

         total time = 2*2.440

                          = 4.88 s