A solid sphere rolls up an inclined plane of inclination angle 35 degrees. At the bottom of the incline, the center of mass of the sphere has a translational speed of 13 ft/s. How long does it take to r etyurn to th bottom?

0.277 s

Explanation:

We first find the height of the incline by considering the potential energy of the solid sphere = kinetic energy of solid sphere + rotational kinetic energy of solid sphere

mgh = 1/2mv² + 1/2Iω²  and ω = v/r where v = speed of centre of mass and r = radius of sphere. I = rotational inertia of solid sphere = 2/5mr²

mgh = 1/2mv² + 1/2I(v/r)² = 1/2mv² + 1/2 × 2/5mr²× v²/r²

mgh = 1/2mv² + 1/5mv² = 7mv²/10

h = 7v²/10g

The torque on the sphere τ = Iα = mgRsinθ, which is the torque of the horizontal component of the weight of the sphere about the centre of mass of the sphere.

2/5mR²α = mgRsinθ

Rα = 5gsinθ/2 = a which is the tangential acceleration of the centre of mass of the sphere.

So, the net acceleration, a₁ moving up the incline is a - gsinθ = 5gsinθ/2 - gsinθ = 3gsinθ/2.

Using s = ut + 1/2at², we find the time it takes the sphere to roll up the incline. s = hsin35 = 7v²sin35/10g = 7 × 13sin35/(10 × 32) = 2.12 ft,  u = v = 13 ft/s and a = 3gsinθ/2 = (3 × 32sin35°)/2 = 27.53 ft/s²

So, s = ut + 1/2at²

2.12 = 13t + 27.53t²/2 = 13t + 13.77t²

13.77t² + 13t - 2.12 = 0

Using the quadratic formula,

t = [-13 ±√(13² - 4 × 13.77 × 2.12)]/(2 × 13.77) = [-13 ±√(169 + 116.89)]/(27.54) = [-13 ±√(285.89)]/(27.54) = [-13 ± 16]/27.54

t = (-13 - 16)/27.54 or (-13 + 16)/27.54

 = -29/27.54 or 3/27.54

 = -1.05 or 0.109 s

We take the positive answer, so it takes 0.109 s to go up the incline.

So, the net acceleration, a₂ moving down the incline is a + gsinθ = 5gsinθ/2 + gsinθ = 7gsinθ/2.

Using v² = u² + 2as, where v is the velocity at the top of the incline,

v² = 13² + 2 × -27.53 × 2.12 = 52.2728 ⇒ v = 7.23 ft/s

Using s = ut + 1/2at², we find the time it takes the sphere to roll up the incline. s = hsin35 = 7v²sin35/10g = 7 × 13sin35/(10 × 32) = 2.12 ft,  u = v = 7.23 ft/s and a = 7gsinθ/2 = (7 × 32sin35°)/2 = 64.24 ft/s²

So, s = ut + 1/2at²

2.12 = 7.23t + 64.24t²/2 = 7.23t + 32.12t²

32.12t² + 7.23t - 2.12 = 0

Using the quadratic formula,

t = [-7.23 ±√(7.23² - 4 × 32.12 × -2.12)]/(2 × 32.12) = [-7.23 ±√(52.2729 + 272.3776)]/(64.24) = [-7.23 ±√(324.6505)]/(64.24) = [-7.23 ± 18.02]/64.24

t = (-7.23 - 18.02)/64.24 or (-7.23 + 18.02)/64.24

 = -25.25/64.24 or 10.79/64.24

 = -0.393 or 0.168 s

We take the positive answer, so it takes 0.168 s  to go down the incline.

So the time it takes to return to the bottom of the incline = time to reach top of incline + time to go down the incline. = 0.109 s + 0.168 s = 0.277 s

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