Question

A regulation basketball has a 25.0-cm diameter and a mass of 0.560 kg. It may be approximated as a thin spherical shell with a moment of inertia (2/3)MR^. Starting from rest, how long will it take a basketball to roll without slipping 4.00 m down an incline at 30.0° to the horizontal?

Answer 1

L = length of the incline = 4 m

= angle of incline = 30

h = height dropped = L Sin = (4) Sin30 = 2 m

M = mass of the basketball = 0.560 kg

R = radius = diameter/2 = 25/2 = 12.5 cm = 0.125 m

v = speed of the ball at the bottom

w = angular speed at the bottom = v/R

Using conservation of energy between top and bottom of the incline

Kinetic energy at bottom = Potential energy at the top

(0.5) M v2 + (0.5) Iw2 = Mgh

M v2 + (2/3) (MR2) (v/R)2 = 2 Mgh

v2 + (2/3) v2 = 2gh

v2 = 6gh/5 Eq-1

Consider the motion along the incline surface

vo = initial speed at the top = 0 m/s

a = acceleration

d = distance traveled = L = 4 m

Using the equation

v2 = vo2 + 2 a d

using Eq-1

6gh/5 = 02 + 2 a (4)

6(9.8)(2)/5 = 02 + 2 a (4)

a = 2.94 m/s2

t = time taken

Using the equation

v = vo + a t

sqrt(6gh/5) = 0 + (2.94) t

sqrt(6(9.8)(2)/5) = 0 + (2.94) t

t = 1.65 sec