Q70P - University Physics with Modern Physics

Question

Question: A solid uniform ball rolls without slipping up a hill (Fig. P10.70). At the top of the hill, it is moving horizontally, and then it goes over the vertical cliff. (a) How far from the foot of the cliff does the ball land, and how fast is it moving just before it lands? (b) Notice that when the balls lands, it has a greater translational speed than when it was at the bottom of the hill. Does this mean that the ball somehow gained energy? Explain!

Step-by-Step Solution

Answer

1. (i) 36.5 m (ii) 28 m/s

Step 1: Translational kinetic energy

The translational kinetic energy is defined as the work required accelerating it from rest to a given velocity of a rigid body.

Step 2: Given Data

\(\begin{array}{l}{v_0} = 25\;{\rm{m/s}}\\h = 28\;{\rm{m}}\end{array}\) 

Step 3(a)(i): Determine how far the ball will go

Conservation of energy 

\(\frac{1}{2}mv_0^2 + \frac{1}{2}I\omega _1^2 = mgh + \frac{1}{2}mv_f^2 + \frac{1}{2}I\omega _f^2 \cdot  \cdot   \cdot  \cdot  \cdot   \cdot \left( 1 \right)\)

Rolling without slipper means \(\omega \;{\rm{ = }}\;\frac{v}{r}\)

\(\begin{array}{c}\frac{1}{2}I{\omega ^2} = \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\left( {\frac{v}{r}} \right)^2}\\ = \frac{1}{5}m{v^2} \cdot  \cdot  \cdot   \cdot  \cdot  \cdot \left( 2 \right)\end{array}\)

Substitute (2) in (1),

\(\begin{array}{c}\frac{1}{2}mv_0^2 + \frac{1}{5}mv_0^2 = mgh + \frac{1}{2}mv_2^2 = \frac{1}{5}mv_2^2\\\frac{7}{{10}}mv_1^2 = mgh + \frac{7}{{10}}mv_2^2\\v_2^2 = \left( {v_1^2 - \frac{{10}}{7}gh} \right)\;\;\;\;\;\;\;\;{\rm{where}}\;{v_1} = 25\;{\rm{m/s}}\\{v_2} = 15.26\;{\rm{m/s}}\end{array}\)

Determine the time in air,

\(\begin{array}{c}{v_{oy}} = 0\\ay = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\y - yo = 28\;{\rm{m/s}}\\{\rm{28  =   }}{{\rm{v}}_{oy}} + \frac{1}{2}ay{t^2}\\{t^2} = 5.71\\t = 2.39\;{\rm{s}}\end{array}\)

 x component distance from cliff

\(\begin{array}{c}d = {v_f}t\\ = 15.26\;{\rm{m/s}} \times 2.39\;{\rm{s}}\\ = 36.47\;{\rm{m}}\\ \simeq 36.5\;{\rm{m}}\end{array}\)

Hence, the ball will go 36.5 m.

Step 3(a)(ii): Determine how fast the will move

\(\begin{array}{c}{v_y} = {v_{oy}} + ayt\\ = 9.8 \times 2.39\\ = 23.42\;{\rm{m/s}}\\{{\rm{v}}_x} = \sqrt {\left( {{{\left( {23.42} \right)}^2} + {{\left( {15.26} \right)}^2}} \right)} \\ = 27.96\;{\rm{m/s}}\end{array}\)

Hence, the speed of the ball is 28 m/s.