Tangential Accn: Particle Starting to Slide Down Smooth Sphere

  • Thread starter Thread starter sruthisupriya
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves a particle sliding down a smooth sphere and seeks to determine its tangential acceleration at the point it breaks off the sphere. The discussion centers around the forces acting on the particle and the conditions for it to leave the surface of the sphere.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of tangential acceleration based on the component of weight and the normal force. There is also a focus on determining the angle at which the particle leaves the sphere, with references to the condition involving the normal reaction force.

Discussion Status

The discussion is active, with participants exploring the conditions for the particle to break off the sphere and questioning the derivation of the angle related to the cosine function. Some participants acknowledge their own confusion regarding the angle calculations, indicating a productive exchange of ideas.

Contextual Notes

There is an emphasis on the need to establish the correct conditions under which the particle leaves the sphere, with specific reference to the normal force becoming zero. Participants are also navigating through potential miscalculations regarding the angles involved.

sruthisupriya
Messages
33
Reaction score
0
A particle starts to slide down a fixed smooth sphere from its top.When it breaks off the sphere,what is its tangential acceleration?
 
Physics news on Phys.org
The tangential acceleration is given by the component of its weight in tangential direction, because the only other force acting on the particle is the normal reaction.

For this first you have to calculate the angle of the radius vector at which the particle leaves the sphere. (cos q = 2/3 ?)

MP
 
Before you start thinking about the tangential acceleration, you will need to figure out when the particle breaks off the sphere. Under what condition does the particle "fall off" the sphere? That is why mukundpa stated that [tex]\cos{q}=\frac{2}{3}[/tex]
 
How did you get [itex]cos \theta = \frac{2}{3}[/itex]? The condition which you speak of is that the reaction of the particle to the surface of the sphere is zero?
 
Doesn't matter, I realized after posting that that I did get [itex]cos \theta = \frac{2}{3}[/itex], its just I was working it out using a different angle, and so kept on getting [itex]sin \theta = \frac{2}{3}[/itex].
 

Similar threads

Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 3 ·
Replies
3
Views
6K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
15
Views
2K
Replies
9
Views
2K
  • · Replies 44 ·
2
Replies
44
Views
7K