Impulse integration for a Tennis Racket hitting a Tennis Ball

AI Thread Summary
Integrating impulse over time from t_i to t_f is necessary because impulse is a function of time, represented as J(t). This approach allows for the calculation of cumulative impulse during a specific time interval. Using J directly without specifying time bounds can lead to confusion, as impulse must be evaluated over time to be meaningful. The discussion clarifies that if impulse values at endpoints are known, they can be used, but otherwise, time integration is essential. Understanding J as a time-dependent function resolves the initial confusion regarding the integration limits.
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this,
1684125887202.png

Can someone please tell me why they integrate the impulse over from ##t_i## to ##t_f##? Why not from ##j_i## to ##j_f##? It seems strange integrating impulse with respect to time.

Many thanks!
 
Physics news on Phys.org
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

For this,
View attachment 326569
Can someone please tell me why they integrate the impulse over from ##t_i## to ##t_f##? Why not from ##j_i## to ##j_f##? It seems strange integrating impulse with respect to time.

Many thanks!
If you are using
##\displaystyle \int d \textbf{J}##
and you have the ##\textbf{J}##'s at the endpoints, the use the ##\textbf{J}##'s.

If you don't have the ##\textbf{J}##'s then you need to use
##\displaystyle \int \textbf{F}(t) \, dt##

-Dan
 
  • Like
Likes member 731016
ChiralSuperfields said:
Can someone please tell me why they integrate the impulse over from ##t_i## to ##t_f##? Why not from ##j_i## to ##j_f##? It seems strange integrating impulse with respect to time.
If you think of ##\vec J## as the cumulative impulse given over some period of time then ##\vec J=\vec J(t)## and it is reasonable to write ##\int_{t=t_i}^{t_f}d\vec J(t)##. But omitting the "t=" from the bounds is a bit naughty.
 
  • Like
Likes topsquark and member 731016
haruspex said:
If you think of ##\vec J## as the cumulative impulse given over some period of time then ##\vec J=\vec J(t)## and it is reasonable to write ##\int_{t=t_i}^{t_f}d\vec J(t)##. But omitting the "t=" from the bounds is a bit naughty.
Thank you for your reply @topsquark and @haruspex!

@haruspex, now that you say J is a function of t I think that helps.

Many thanks!
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top