Max Tangential Speed of Mass Before Cord Breaks

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Homework Help Overview

The discussion revolves around a physics problem involving the maximum tangential speed of a mass attached to a string, focusing on tension in both horizontal and vertical circular motion contexts.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply formulas related to force, acceleration, and tension to solve for maximum speed and tension in different scenarios. Some participants question the need to consider gravitational forces in the vertical motion case.

Discussion Status

Participants are exploring the implications of gravity on tension in vertical motion. Some guidance has been provided regarding the need to account for gravitational forces, but there is no explicit consensus on the correct approach yet.

Contextual Notes

The original poster mentions using specific formulas for both parts of the problem, but indicates difficulty in achieving the correct answer for part b. The discussion includes a request for a diagram, suggesting that visual representation may be relevant to understanding the problem setup.

MutTurwen
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A mass M of 2.47 kg is attached to the end of a string whose length is 0.410 m. The mass slides without friction on a horizontal surface as indicated in the diagram.

a)If the string can withstand a maximum tension of 113.1 N, what is the maximum tangential speed the ball can have before the cord breaks?
Answ: 4.33 m/s

b)The mass, with a new string, is whirled in a vertical circle of the same radius about a fixed point. Find the magnitude of the tension when the mass is at the top if its speed at the top is 6.69 m/s.
Answ: 2.45×102 N


For part a) i used the formulaa 1) F=ma 2)a=rw^2 3) V=rw to get the right answer but for part b) I used 1)V=rw 2) a=rw^2 3)T=ma but I am not getting the right answer. Am I doing something wrong?
 
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Diagram please.

BJ
 
b) the motion is vertical, so you must take gravity into account.
 
In the second case, gravity is also acting.
So, mv^2/r=mg+T.
 
T = m ((v^2/r)-g)
 

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