A 160 g basketball has a 32.7 cm diameter and may be approximated as a thin spherical shell. Starting from rest, how long will it take a basketball to roll without slipping 3.16 m down an incline that makes an angle of 27◦ with the horizontal? The moment of inertia of a thin spherical shell of radius R and mass m is I = 2 3 m R2 , the acceleration due to gravity is 9.8 m/s 2 , and the coefficient of friction is 0.3 . Answer in units of s.

As seen in the attached diagram, we are going to use dynamics to resolve the problem, so we will be using the equations for the translation and the rotation dyamics:

Translation: ΣF = ma

Rotation: ΣM = Iα ; where α = angular acceleration

Because the angular acceleration is equal to the linear acceleration divided by the radius, the rotation equation also can be represented like:

ΣM = I(a/R)

Now we are going to resolve and combine these equations.

For translation: Fx - Ffr = ma

We know that Fx = mgSin27°, so we substitute:

(1) mgSin27° - Ffr = ma

For rotation: (Ffr)(R) = (2/3mR²)(a/R)

The radius cancel each other:

(2) Ffr = 2/3 ma

We substitute equation (2) in equation (1):

mgSin27° - 2/3 ma = ma

mgSin27° = ma + 2/3 ma

The mass gets cancelled:

gSin27° = 5/3 a

a = (3/5)(gSin27°)

a = (3/5)(9.8 m/s²(Sin27°))

a = 2.67 m/s²

If we assume that the acceleration is a constant we can use the next equation to find the velocity:

V = √2ad; where d = 0.327m

V = √2(2.67 m/s²)(0.327m)

V = 1.32 m/s

Because V = d/t

t = d/V

t = 0.327m/1.32 m/s

t = 0.24 s