Question

A regulation basketball has a 19 cm diameter

and may be approximated as a thin spherical

shell.

How long will it take a basketball starting

from rest to roll without slipping 4.0 m down

an incline that makes an angle of 45.4" with

the horizontal? The acceleration of gravity is

9.81 m/s.

Answer in units of s.

Answer 1

1.27 sec

Explanation:

From the given information:

For the ball, Moment of inertia
`I = (2)/(5)mr^2`

When the height h of the ball is:

h = 4.0 sin 45.4°

and the initial potential energy = mgh

⇒ mg(4.0 sin 45.4°)

According to the conservation of energy.

`mgh = (1)/(2)mv^2 + (1)/(2)I \omega^2`

`mgh = (1)/(2)mv^2 + (1)/(2) ( (2)/(5)mr^2) \omega^2`

`mgh = (1)/(2)mv^2 + (1)/(2) ( (2)/(5)mr^2) (v^2)/(r^2 )`

`gh = (1)/(2)v^2 + (1)/(2) ( (2)/(5)) v^2`

`gh = 0.5v^2 + 0.5( 0.4 )v^2`

`gh = 0.5v^2 + 0.2v^2`

`gh = 0.7v^2`

∴

`(9.81)(4.0 sin 45.4^0) = 0.7 v^2`

`27.94 = 0.7 v^2`

`v^2= (27.94)/(0.7)`

`v^2=39.91`

`v=√(39.91)`

v = 6.32 m/s

`t= (2s)/(v)`

`t =(2* 4.0)/(6.32)`

t = 1.27 sec