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October 27th, 2019, 06:14 AM  #1 
Newbie Joined: Aug 2019 From: India Posts: 23 Thanks: 1  Why wouldn't the horizontal component of the force make yoyo to move ?
The problem statement is: A yoyo rests on a table (see the attachment) and the free end of its string is gently pulled at an angle $\displaystyle \theta$ to the horizontal as shown. What is the critical value of $\displaystyle \theta$ such that the yoyo remains stationary, even though it is free to roll? (This problem may be solved geometrically if you consider the torques about P, the point of contact with the table). Now, the thing is that this problem can be solved ( I mean answer can be made to come right) if we assume a few things and those assumptions are my problem. The proposed solution is: (see the second attachment for reference) The angle between $\displaystyle r$ and $\displaystyle R$ is $\displaystyle \theta$ because of the principle the angle between two lines is the angle between their normals and $\displaystyle r$ is perpendicular to the line of force because it is tangent to that inner circle and $\displaystyle R$ is perpendicular to horizontal line. Now just equate the torques produced by these forces: $\displaystyle F~\sin\theta ~r \sin\theta = F~\cos\theta~(Rr\cos\theta)$ $\displaystyle r \sin^2\theta = R~\cos\theta  r~\cos^2 \theta$ $\displaystyle r= R~\cos\theta$ $\displaystyle \theta = \cos^{1} (r/R)$ (I have included the solution with all possible intricacies only because to save your precious time, I know these things are so easy for you and my intention was not to exhibit anything it was purely for clarity and time saving.) Now, how can we ever ignore the translational effect of $\displaystyle F~cos\theta$? I mean it would cause the the yoyo to translate in horizontal direction if there were no friction. And if there were to have a friction then for critical angle our equation would be very simple $\displaystyle F~\cos\theta + Friction =0$ $\displaystyle \theta = \cos^{1} (Friction/F)$. So why we are not considering it? I mean why are we just assuming that $\displaystyle \Sigma F=0$ and $\displaystyle \Sigma \tau=0$ (I must clarify that some people in US call torque as moment and denote it by M, here moment has been phrased as torque and is represented by tau \tau). What are the reasons for taking those two assumptions? Thank you. Any help will be much appreciated. Last edited by skipjack; October 27th, 2019 at 09:15 AM. 
October 27th, 2019, 08:12 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 
Reference the diagram ... As shown above, all forces acting on the yoyo (string tension F , weight mg, normal force N, and static friction f) act through point P, the point of contact with the ground, hence the net torque on the yoyo is zero. As you stated, the angle $\theta$ where the yoyo remains in equilibrium is $\theta = \arccos\left(\dfrac{r}{R}\right)$ Reviewing the equilibrium of the translational forces ... $F\cos{\theta} = f$ $N + F\sin{\theta} = mg$ ... and the rotational forces about point C (the center) $F \cdot r  f \cdot R = 0$ $F \cdot r  F\cos{\theta} \cdot R = 0$ $F(r  \cos{\theta} \cdot R) = 0$ $F \ne 0 \implies r = R\cos{\theta} \implies \cos{\theta} = \dfrac{r}{R}$ 
October 27th, 2019, 08:38 AM  #3 
Newbie Joined: Aug 2019 From: India Posts: 23 Thanks: 1 
I have one doubt : the tension force is applied at the inner circle so we should find the effective lever arm from P to that point. But you have made that as if string force is acting on P. I mean that tension force is in contact with the inner circle so I think it would cause a net torque at P. I request you to please clear my doubt

October 27th, 2019, 08:53 AM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677  Quote:
Is not the lever arm relative to the hinge zero? Last edited by skeeter; October 27th, 2019 at 09:43 AM.  
October 27th, 2019, 09:56 AM  #5 
Newbie Joined: Aug 2019 From: India Posts: 23 Thanks: 1 
Thank you so much for being so clear.


Tags 
component, equilibrium, force, horizontal, make, move, yoyo 
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