What is the Approximate Torque on a Baseball Bat Held Horizontally?

[disque]

Homework Statement

A baseball player holds a bat at the end of the handle. If the bat is held horizontally, what is the approximate torque due to the force of gravity on the bat, with a pivot point at the batter's hands? Assume that the the bat is uniform. Use L for length of the bat, m for mass and g for acceleration due to gravity.

Homework Equations

t=d(lever arm)F

The Attempt at a Solution

I thought it was just lmg and that was wrong so I tried lm-g and that was wrong as well so I don't know what I'm doing wrong. Any help would be appreciated!

 

[disque]

I got it (lmg)/2 :)

 

[nlightner]

I would like to clarify that torque is a measure of the force that causes an object to rotate around an axis. In this case, the axis of rotation is at the batter's hands. The formula for torque is t = r x F, where r is the distance from the axis of rotation to the point where the force is applied, and F is the magnitude of the force.

In this scenario, the force of gravity is acting on the bat, and the point of application is at the center of mass of the bat. Therefore, the distance from the axis of rotation to the point of application is half the length of the bat (L/2).

The formula for torque can be rewritten as t = r x mg, where r = L/2 and mg is the weight of the bat. Substituting these values, we get t = (L/2) x mg = (Lmg)/2. This is the approximate torque due to the force of gravity on the bat, with a pivot point at the batter's hands.

I would also like to mention that the units for torque are Newton-meters (Nm). Therefore, the final answer would be in Nm units. I hope this helps clarify the concept of torque and how it applies to this scenario.