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

Answer:

1.27 sec

Step-by-step explanation:

From the given information:

For the ball, Moment of inertia $I = \dfrac{2}{5}mr^2$

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.

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I \omega^2$

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2} ( \dfrac{2}{5}mr^2) \omega^2$

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2} ( \dfrac{2}{5}mr^2) \dfrac{v^2}{r^2 }$

$gh = \dfrac{1}{2}v^2 + \dfrac{1}{2} ( \dfrac{2}{5}) v^2$

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

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

$gh = 0.7v^2$

∴

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

$27.94 = 0.7 v^2$

$v^2= \dfrac{27.94}{0.7}$

$v^2=39.91$

$v=\sqrt{39.91}$

v = 6.32 m/s

$t= \dfrac{2s}{v}$

$t =\dfrac{2* 4.0}{6.32}$

t = 1.27  sec