Question

1. A bowling ball of mass 7.72kg and radius 9.30cm rolls without slipping down a lane at 3.78m/s. Calculate its total kinetic energy.

I need the correct answer in order to give full rating. thanks.

Answer 1

the condition for rolling without slipping is $v=r\omega$

$\omega =\frac{v}{r}$        =40.64 rad/s

total kinetic energy is $k=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega ^{2}$

moment of inertia of a solid ball is $\frac{2}{5}mr^{2}$

there fore on substituting we get $k=\frac{7}{10}mv^{2}$     =   77.21 J

Answer 2

The total kinetic energy of the bowling ball is,

               E = (1/2)mv² + (1/2)Iw²

Here, I = (2/5)mr², and w = v/r. Substitute these relations in above equation

we get the following equation.

             E = (7/10)mv²

                 = (7/10)(7.72 kg)(3.78 m/s)²

                 = 77.2 kg