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

From the given data,

For the spherical shell is
`$(k^2)/(R^2)= (2)/(3)$`

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

Therefore,

`$a =(g \sin \theta)/(1+ (k^2)/(R^2)) $`

`$= (g \sin \theta)/(1 +(2)/(3))$`

`$= (3)/(5) g \sin \theta$`

= 0.6 x 9.81 x sin ( 29.7)

`$= 2.913 \ m/s^2$`

`$h = (1)/(2) a(\Delta t)^2$`

`$\Delta t = \sqrt{(2h)/(a)}$`

`$\Delta t = \sqrt{(2 * 2.9)/(2.913)}$`

Δt = 1.411 s