Question

A regulation basketball has a 15 cm diameter and may be appropriated as a thin spherical shell.

How long will it take a basketball starting from rest to roll without slipping 2.9 m down an incline that makes an angle of 29.7 with the horizontal? The acceleration of gravity is 9.81 m/s^2 .

Answer in units of s.

Answer 1

Solution :

From the given data,

For the spherical shell is $\frac{k^2}{R^2}= \frac{2}{3}$

where x is the radius of gyration and the acceleration of a rolling body on an inclined plane = a

Therefore,

$a =\frac{g \sin \theta}{1+ \frac{k^2}{R^2}}$

  $= \frac{g \sin \theta}{1 +\frac{2}{3}}$

  $= \frac{3}{5} g \sin \theta$

  = 0.6 x 9.81 x sin ( 29.7)

  $= 2.913 \ m/s^2$

$h = \frac{1}{2} a(\Delta t)^2$

$\Delta t = \sqrt{\frac{2h}{a}}$

$\Delta t = \sqrt{\frac{2 \times 2.9}{2.913}}$

Δt = 1.411 s