Ball hitting baseball bat (with no impulse)

Click For Summary
SUMMARY

The discussion revolves around calculating the distance x at which a baseball should be positioned to ensure no impulse reaction occurs at point A when a bat strikes the ball. The solution employs the principles of conservation of linear and angular momentum, specifically using the equations M(vg1) + ∑ Fxdt = M(vg2) and Igω1 + ∑ Madt = Igω2 + mux. The user identifies a potential error in their approach, questioning whether the inertia was incorrectly applied for the center of gravity (COG) of the bat instead of the end point. The correct formula derived for x is L/(2(1-mu)).

PREREQUISITES
  • Understanding of conservation of linear momentum
  • Familiarity with conservation of angular momentum
  • Knowledge of moment of inertia calculations
  • Basic principles of dynamics and kinematics
NEXT STEPS
  • Study the application of conservation laws in collision problems
  • Learn about moment of inertia for different shapes and axes
  • Explore angular momentum in rigid body dynamics
  • Investigate impulse-momentum theorem and its applications
USEFUL FOR

Students and professionals in physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to momentum and collisions.

SnappySeudonym
Messages
1
Reaction score
0

Homework Statement




Given the setup above, what distance x should the ball be away so that there is no impulse reaction at A?

Homework Equations


Conservation of Linear and Angular momentum.

The Attempt at a Solution


Conservation of linear momentum (←+)
(considering the bat alone)
M(vg1) + ∑ Fxdt = M(vg2)
M(vg2 - (vg1)) = ∑ Fxdt
(3/2)LM(ω2 - ω1) = ∑ Fxdt ⇒ (1)

Conservation of angular momentum (+⊃)
Igω1 + ∑ Madt = Igω2 +mux
Ig = (1/12)M(3L)2
Ig = (3/4)L2M
(3/4)L21 + ∑ (Fxdt)x = (3/4)L22 + mux
using (1)
(3/4)L21 +((3/2)LM(ω2 - ω1))x = (3/4)L22 + mux
x((3/2)LM(ω2 - ω1 - mux) = (3/4)L2M(ω2 - ω1)
x = L/(2(1-mu)

I've seen this problem before, but not with the bat/rod moving at an initial angular speed, can anyone give some insight as to where my solution is wrong, Is it because a used inertia for the COG of the bar and not the end?
 

Attachments

  • FIG.JPG
    FIG.JPG
    12.2 KB · Views: 663
Physics news on Phys.org
SnappySeudonym said:
Conservation of angular momentum (+⊃)
Ig = (1/12)M(3L)2
:
∑ (Fxdt)x
The first equation above appears to be taking moments about the centre of the rod (1/12), but the second expression is a moment about the suspension point (x).
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
Replies
10
Views
3K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 4 ·
Replies
4
Views
961
Replies
14
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 6 ·
Replies
6
Views
7K