Question

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.

Answer 1

Answer:

   t = 0.24 s

Explanation:

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