A batting cage is designed by gluing a baseball to a spring-mass-dashpot system so that the ball is returned to the batter. The weight has mass 0.5kg which is attached to a spring which is stretched 2 meters by a force of 8 Newtons. Suppose the ball starts at equilibrium and a baseball player hits the ball, launching it at 20 meters per second. How strong should the dashpot be (in Newtons per meter/second i.e. the value of ”c” in class) so that when the ball returns to the hitter, it is traveling at a speed of 2 meters per second. [Hint: overdamping would prohibit the ball from ever returning to the hitter.]
To find the strength of the dashpot in Newtons per meter/second, we need to use the concept of damped harmonic motion.
Damped harmonic motion refers to the motion of an object in a system where there is resistance (damping) to its motion. In this case, the dashpot provides the damping force.
The equation for damped harmonic motion is given by:
m * d^2x/dt^2 + c * dx/dt + k * x = 0
Where:
- m is the mass of the object (0.5 kg)
- c is the damping coefficient (strength of the dashpot in Newtons per meter/second)
- k is the spring constant (force required to stretch the spring 1 meter, which is 8 Newtons/2 meters = 4 N/m)
- x is the displacement of the object from equilibrium (position of the ball)
Since the system is designed in such a way that the ball returns to the batter, we know that the ball has to come to rest (velocity is zero) at the equilibrium position. Therefore, we need to find the value of c, such that when the ball returns to the hitter, it is traveling at a speed of 2 meters per second (v = 2 m/s).
To solve this problem, we can use the initial conditions (when the ball is hit) and the final conditions (when the ball returns to the hitter at a speed of 2 m/s) to solve for c.
Let's use the initial condition first. When the ball is hit, its velocity v is given as 20 m/s. Therefore:
v = dx/dt
20 = dx/dt (Equation 1)
Now, let's use the final condition. When the ball returns to the hitter, its velocity v is given as 2 m/s. Therefore:
v = dx/dt
2 = dx/dt (Equation 2)
We need to solve Equations 1 and 2 along with the equation of motion to find the value of c.
m * d^2x/dt^2 + c * dx/dt + k * x = 0
Substituting the initial condition into the equation of motion:
(0.5) * d^2x/dt^2 + c * 20 + (4) * x = 0
0.5 * d^2x/dt^2 + 4 * x = -20c (Equation 3)
Substituting the final condition into the equation of motion:
(0.5) * d^2x/dt^2 + c * 2 + (4) * x = 0
0.5 * d^2x/dt^2 + 4 * x = -2c (Equation 4)
Now, we need to subtract Equation 4 from Equation 3 to eliminate the c term:
(-20c) - (-2c) = 0
-20c + 2c = 0
-18c = 0
From this, we can see that c must be equal to zero, which implies that there should be no damping (dashpot should have no strength).
Therefore, in order for the ball to return to the hitter at a speed of 2 m/s, there should be no damping in the system (no strength in the dashpot).