Question

A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 9.37 m/s at the bottom of the rise. Find the translational speed at the top.

Answer 1

Given is:-

Vertical rise = 0.760 m

translational speed of the ball  

Now,

By applying conservation of energy we get

$\frac{1}{2}mu^2+ \frac{1}{2}Iw_i^2 = mgh + \frac{1}{2}mv^2 + \frac{1}{2}Iw_f^2$

by plugging moment of inertia of sphere $I = \frac{2}{5}mr^2$ and cancelling mass m from both sides we get

$\frac{1}{2}u^2+ \frac{1}{2}(\frac{2}{5}r^2)w_i^2 = gh + \frac{1}{2}v^2 + \frac{1}{2}(\frac{2}{5}r^2)w_f^2$

by plugging all the values we get

$\frac{1}{2}(9.37 m/s)^2+ \frac{1}{2}(\frac{2}{5}r^2)w_i^2 = (9.8 m/s^2)(0.760m)+ \frac{1}{2}v^2 + \frac{1}{2}(\frac{2}{5}r^2)w_f^2$

or

$\frac{1}{2}(9.37 m/s)^2+ \frac{1}{2} \times \frac{2}{5}u^2 = (9.8 m/s^2)(0.760m)+ \frac{1}{2}v^2 + \frac{1}{2} \times\frac{2}{5}v^2$

thus

$\frac{1}{2}(9.37 m/s)^2+ \frac{1}{2} \times \frac{2}{5}(9.37 m/s)^2 = (9.8 m/s^2)(0.760m)+ \frac{1}{2}v^2 + \frac{1}{2} \times\frac{2}{5}v^2$

which gives us the final translational speed at the top , which is

$\boxed{v = 8.7839 m/s}$