A baseball bat has a sweet spot where a ball can be hit with almost effortless transmission of energy. A careful analysis of baseball dynamics shows that this special spot is located at the point where an applied force would result in pure rotation of the bat about the handle grip. Determine the location of the sweet spot, xₛ, of the bat shown in Fig. 11–53. The linear mass density of the bat is given roughly by (0.61 + 3.3x²) kg/m, where x is in meters measured from the end of the handle. The entire bat is 0.84 m long. The desired rotation point should be 5.0 cm from the thin end where the bat is held. [Hint: Where is the cm of the bat?]

Verified step by step guidance

1. Step 1: Understand the problem. The sweet spot is the point where an applied force results in pure rotation about the handle grip. To find this, we need to calculate the center of mass (cm) of the bat and the moment of inertia about the desired rotation point. The linear mass density is given as \( \lambda(x) = 0.61 + 3.3x^2 \) (in kg/m), where \( x \) is the distance from the handle end. The bat is 0.84 m long, and the rotation point is 5.0 cm (0.05 m) from the handle end.
2. Step 2: Calculate the total mass of the bat. The total mass \( M \) is the integral of the linear mass density over the length of the bat: \( M = \int_0^{0.84} \lambda(x) dx = \int_0^{0.84} (0.61 + 3.3x^2) dx \). Perform the integration to express \( M \) in terms of the given limits.
3. Step 3: Determine the center of mass (cm) of the bat. The center of mass \( x_{cm} \) is given by \( x_{cm} = \frac{1}{M} \int_0^{0.84} x \lambda(x) dx \). Substitute \( \lambda(x) = 0.61 + 3.3x^2 \) and integrate \( \int_0^{0.84} x(0.61 + 3.3x^2) dx \) to find \( x_{cm} \).
4. Step 4: Calculate the moment of inertia about the desired rotation point. The moment of inertia \( I \) is given by \( I = \int_0^{0.84} \lambda(x) (x - 0.05)^2 dx \). Substitute \( \lambda(x) = 0.61 + 3.3x^2 \) and integrate \( \int_0^{0.84} (0.61 + 3.3x^2)(x - 0.05)^2 dx \) to find \( I \).
5. Step 5: Use the physical condition for the sweet spot. The sweet spot is located where the applied force results in pure rotation about the handle grip. This occurs when the torque due to the center of mass balances the rotational inertia. Solve for \( x_s \) using the relationship between the center of mass, moment of inertia, and the desired rotation point.

Key Concepts

Center of Mass

The center of mass (cm) of an object is the point where its mass is evenly distributed in all directions. For a baseball bat, the cm can be calculated by integrating the mass distribution along its length. Understanding the location of the cm is crucial for determining how forces applied to the bat will affect its motion, particularly in rotational dynamics.

Torque

Torque is a measure of the rotational force applied to an object, calculated as the product of the force and the distance from the pivot point (lever arm). In the context of the baseball bat, the sweet spot is where the applied force results in pure rotation, meaning that the torque about the handle grip is maximized. This concept is essential for analyzing how the bat will rotate when a ball is hit.

Linear Mass Density

Linear mass density is defined as the mass per unit length of an object, which varies along the length of the bat in this case. The given function (0.61 + 3.3x²) kg/m indicates that the mass distribution changes with position, affecting both the center of mass and the moment of inertia. Understanding linear mass density is vital for calculating the bat's dynamics and locating the sweet spot.