Rotation qns which i am stuck for 2 hours

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Homework Help Overview

The discussion revolves around a physics problem involving rotational motion and friction, specifically focusing on a gymnast tossing a hoop with a backward spin. The problem requires determining the minimum backward angular velocity needed for the hoop to eventually spin back after being thrown forward.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between torque, angular momentum, and kinetic energy in the context of the hoop's motion. Questions arise about the application of conservation laws and the effects of friction on the system.

Discussion Status

Some participants have offered insights into the conservation of angular momentum when considering the torque about the contact point. There is an ongoing exploration of the relationship between translational and rotational kinetic energy, with references to conditions under which the hoop may roll backward.

Contextual Notes

The problem includes considerations of both static and kinetic coefficients of friction, which may influence the dynamics of the hoop's motion. Participants are navigating the implications of these factors on the solution process.

Wen
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I have been stuck at this qns for 2 hours.

A gymnast often toss a hoop forward while giving it a backward spin. the hoop has a radius R and mass M, and is thrown forward with speed V. Find the minimum backward angular velocity w the gymnast must give in order to make sure it spin back eventually . Take both its static and kinectic coefficient of friction to be #.

Since friction is acting , its nett torque is not 0. Hence conserv. of angular momentum cannot be applied.

if its nett torque is Mg#R, how should i relate this to the translational and rotational K.E.?

Please teach me how to solve this.
 
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if the torque is taken about contact point(bottum) then torque due to all three forces is zero and we can conserve angular momentum.
 
so 1/2 I w^2= 1/2 MV^2 ?
 
yes, the kinetic energy of a rolling object is, as you have stated:
[tex]KE_{rolling} = \frac{1}{2}mv^2 + \frac{1}{2}I{\omega}^2[/tex]

if the object is on the veeeeeeerge of rolling backwards, or back to the thrower, the inequality you have stated works.

Regards,

Nenad
 

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