Question

A bowling ball encounters a 0.760 m vertical rise on the way back to the ball rack. Ignore frictional losses and assume the mass of the ball is distributed uniformly. the translational speed pf the ball is 3.5 m/s at the bottom of the rise. find the translational speed at the top. note that moment of inertia of a solid sphere about an axis tanget to its surface is I = (7/5) MR^2

Answer 1

To solve this exercise we need the concept of Kinetic Energy and its respective change: Initial and final kinetic energy.

Let's start considering that the angular velocity is given by,

$\omega = \frac{v}{R}$

Where,

V = linear speed

R = the radius

In the case of the initial kinetic energy:

$KE_i=\frac{1}{2} mv^2 + \frac{1}{2}I \omega^2$

Where I is the moment of inertia previously defined.

$KE_i = \frac{1}{2}(m)3.5^2 + \frac{1}{2}* (\frac{2}{5} m R^2) (\frac{3.5}{R})^2$

In the case of the final kinetic energy, we have to,

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

$KE_f = m * 9.81 * 0.76 + \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2$

For conservation of Energy we have, that

$KE_f = KE_i$, then (canceling the mass and the radius)

$\frac{1}{2} 3.5^2 + \frac{1}{2}(\frac{2}{5})(3.5)^2= 9.81 * 0.76 + \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

$8.575= 7.4556+ \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

$1.1194= \frac{1}{2}( v^2 + (\frac{2}{5}) (v)^2)$

$2.2388= (\frac{7}{5}) (v)^2$

$v=1.26m/s$