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 4.43 m/s at the bottom of the rise. Find the translational speed at the top.
3.96m/s
To find the translational speed at the top of the vertical rise, we can use the principle of conservation of mechanical energy. This principle states that the total mechanical energy of an object is conserved throughout its motion if no external work is done on it.
The total mechanical energy is the sum of the object's kinetic energy (KE) and potential energy (PE). At the bottom of the rise, all of the ball's initial kinetic energy is in the form of translational kinetic energy. At the top of the rise, the ball has reached its maximum height and has zero translational kinetic energy. Therefore, its kinetic energy at the top is zero.
We can write the conservation of mechanical energy equation as follows:
KE_bottom + PE_bottom = KE_top + PE_top
Since the ball has uniform mass distribution, its potential energy is given by the equation:
PE = m * g * h
where m is the mass of the ball, g is the acceleration due to gravity, and h is the height.
Let's denote the height of the rise as h_rise.
At the bottom of the rise:
KE_bottom = (1/2) * m * v_bottom^2
At the top of the rise:
KE_top = 0
PE_top = m * g * h_rise
Rearranging the conservation of mechanical energy equation, we get:
KE_bottom + PE_bottom = KE_top + PE_top
(1/2) * m * v_bottom^2 + m * g * h_rise = 0 + m * g * h_rise
Cancelling out the mass from both sides and rearranging the equation, we have:
(1/2) * v_bottom^2 = g * h_rise
Now, we can solve for the translational speed (v_top) at the top of the rise:
v_top = sqrt(2 * g * h_rise)
Plugging in the values, with g being the acceleration due to gravity (approximated as 9.8 m/s^2) and h_rise being the vertical rise height (0.76 m), we can calculate the translational speed at the top:
v_top = sqrt(2 * 9.8 * 0.76)
v_top ≈ 3.20 m/s
Therefore, the translational speed of the bowling ball at the top of the vertical rise is approximately 3.20 m/s.
To find the translational speed of the bowling ball at the top of the vertical rise, we can apply the principle of conservation of mechanical energy. According to this principle, the total mechanical energy of the system remains constant, provided there is no external work acting on the system.
The mechanical energy of the ball can be divided into two components: kinetic energy (KE) and potential energy (PE). At the bottom of the rise, the ball only possesses kinetic energy since its height is zero.
Using the equations for kinetic energy (KE = 1/2 * mv^2) and potential energy (PE = mgh), where m is the mass of the ball, v is the translational speed, g is the acceleration due to gravity, and h is the height, we can set up the following equation:
KE_initial + PE_initial = KE_final + PE_final
At the bottom of the rise, the height is zero, so the potential energy is also zero. The equation becomes:
0.5mv_initial^2 = 0.5mv_final^2 + mgh
Here, we know the initial translational speed (v_initial = 4.43 m/s) and the height of the rise (h = 0.760 m). We need to find v_final, the translational speed at the top.
Plugging in the values into the equation, we get:
0.5 * m * (4.43 m/s)^2 = 0.5 * m * v_final^2 + m * g * 0.760 m
Simplifying:
0.5 * (4.43 m/s)^2 = 0.5 * v_final^2 + 9.8 m/s^2 * 0.760 m
19.595 m^2/s^2 = 0.5 * v_final^2 + 7.448 m^2/s^2
Rearranging the equation:
0.5 * v_final^2 = 19.595 m^2/s^2 - 7.448 m^2/s^2
0.5 * v_final^2 = 12.147 m^2/s^2
Dividing by 0.5:
v_final^2 = 24.294 m^2/s^2
Taking the square root of both sides:
v_final = √(24.294) m/s
v_final ≈ 4.929 m/s
Therefore, the translational speed of the bowling ball at the top of the vertical rise is approximately 4.929 m/s.