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 5.40 m/s at the bottom of the rise. Find the translational speed at the top.
To find the translational speed of the ball at the top of the rise, we can use the principle of conservation of mechanical energy. This principle states that the total mechanical energy of an object is conserved as long as there are no external forces doing work on the object.
In this case, the mechanical energy of the ball consists of its kinetic energy (due to its translational speed) and its gravitational potential energy (due to its height from the ground). We can write the equation for conservation of mechanical energy as:
KE_bottom + PE_bottom = KE_top + PE_top
The kinetic energy at the bottom of the rise is given by:
KE_bottom = (1/2) * m * v_bottom^2
where m is the mass of the ball and v_bottom is the translational speed at the bottom of the rise.
The potential energy at the bottom of the rise is zero, since the height is measured relative to the ground.
The potential energy at the top of the rise is given by:
PE_top = m * g * h
where h is the height of the rise and g is the acceleration due to gravity.
Now, we can rearrange the equation for conservation of mechanical energy to solve for the translational speed at the top:
KE_top = KE_bottom + PE_bottom - PE_top
Substituting the expressions for kinetic and potential energy, we get:
(1/2) * m * v_top^2 = (1/2) * m * v_bottom^2 + 0 - m * g * h
Canceling out the mass, we have:
(1/2) * v_top^2 = (1/2) * v_bottom^2 - g * h
Finally, we can solve for the translational speed at the top:
v_top = sqrt(v_bottom^2 - 2 * g * h)
Substituting the given values:
v_top = sqrt((5.40 m/s)^2 - 2 * 9.8 m/s^2 * 0.760 m)
Calculating the expression inside the square root:
v_top = sqrt(29.16 m^2/s^2 - 14.98 m^2/s^2)
v_top = sqrt(14.18 m^2/s^2)
v_top ≈ 3.77 m/s
Therefore, the translational speed of the ball at the top of the rise is approximately 3.77 m/s.