Hoop rolling up an inclined plane A hollow cylinder (or hoop) is rolling along a horizontal surface with speed v = 3.3 m/s when it reaches a 15◦ incline. (a) How far up the incline will it go? (b) How long (in seconds) will it be on the incline before it arrives back at the bottom?

For rolling up or down an incline plane, the acceleration or deceleration of the rolling body is given by the following expression

a = g sinθ / (1 + k²/r²)

where k is radius of gyration of rolling body and θ is angle of inclination

a = g sin15 / (1 + 1) [ for hoop k = r ]

a = 9.8 x. 2588 / 2

= 1.268 m / s²

a)

Let s be the distance up to which it goes

v² = u² - 2as

0 = 3.3² - 2 x 1.268 s

s = 4.3 m

b) Let time in going up be t₁

v = u - at₁

0 = 3.3 - 1.268 t₁

t₁ = 2.6 s

Time in going down t₂

s = 1/2 a t₂²

4.3 =.5 x 1.268 t₂²

t₂ = 2.60

Total time

= t₁ + t₂

= 2.6 + 2.6

= 5.2 s