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?

Explanation:

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

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force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2