Find the minimum speed of a yo-yo, revolving in a vertical circle, so that the cord does not slacken

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A yo-yo is swung with a constant speed in a vertical circle. If the yo-yo has a mass of 80 g and the radius of the circle is 1.5 m, find the minimum speed that this yo-yo must have at the top of the circle so that the cord does not slacken.






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    A yo-yo is swung with a constant speed in a vertical circle. If the yo-yo has a mass of 80 g and the radius of the circle is 1.5 m, find the minimum speed that this yo-yo must have at the top of the circle so that the cord does not slacken.






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      A yo-yo is swung with a constant speed in a vertical circle. If the yo-yo has a mass of 80 g and the radius of the circle is 1.5 m, find the minimum speed that this yo-yo must have at the top of the circle so that the cord does not slacken.






      homework-and-exercises newtonian-mechanics







      share|cite|improve this question















      A yo-yo is swung with a constant speed in a vertical circle. If the yo-yo has a mass of 80 g and the radius of the circle is 1.5 m, find the minimum speed that this yo-yo must have at the top of the circle so that the cord does not slacken.






      homework-and-exercises newtonian-mechanics




      homework-and-exercises newtonian-mechanics






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      edited 1 hour ago









      Qmechanic♦

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      asked 3 hours ago









      Archipelago2000

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          If the Yo-Yo string doesn't slacken, the centrifugal force at the top is greater than or equal to the force exerted due to the weight of the Yo-Yo (mass of string is considered negligible here). Only then there is some tension in the string to prevent it from slackening.



          beginalignrequirecancelcancelm,fracv^2r &ge cancelmg\ fracv^2r &ge g\ v &gesqrtgrendalign



          $m$: mass of the Yo-Yo, $v$: velocity at highest point,



          $r$: vertical radius of the circle (or length of the string) and $g$ is acceleration due to gravity.



          Since you are looking for minimum speed, we can take minimum value of $v$ , that is



          $sqrtgr=sqrt(9.8times1.5)=3.83largefracmathrmmmathrmsec$



          which is basically independent of the mass of the yo-yo.






          share|cite|improve this answer






















          • @santimirandarp is not not possible to get 'greater than or equal to' symbol up there in the first three equation?
            – user64829
            2 hours ago










          • Is it better now? You can edit it by yourself by analogy with the symbols I've used!
            – santimirandarp
            1 hour ago










          • Yeah, much better. Sorry for the trouble, I'm writing here for the first time.
            – user64829
            47 mins ago










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          1 Answer
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          up vote
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          If the Yo-Yo string doesn't slacken, the centrifugal force at the top is greater than or equal to the force exerted due to the weight of the Yo-Yo (mass of string is considered negligible here). Only then there is some tension in the string to prevent it from slackening.



          beginalignrequirecancelcancelm,fracv^2r &ge cancelmg\ fracv^2r &ge g\ v &gesqrtgrendalign



          $m$: mass of the Yo-Yo, $v$: velocity at highest point,



          $r$: vertical radius of the circle (or length of the string) and $g$ is acceleration due to gravity.



          Since you are looking for minimum speed, we can take minimum value of $v$ , that is



          $sqrtgr=sqrt(9.8times1.5)=3.83largefracmathrmmmathrmsec$



          which is basically independent of the mass of the yo-yo.






          share|cite|improve this answer






















          • @santimirandarp is not not possible to get 'greater than or equal to' symbol up there in the first three equation?
            – user64829
            2 hours ago










          • Is it better now? You can edit it by yourself by analogy with the symbols I've used!
            – santimirandarp
            1 hour ago










          • Yeah, much better. Sorry for the trouble, I'm writing here for the first time.
            – user64829
            47 mins ago














          up vote
          3
          down vote













          If the Yo-Yo string doesn't slacken, the centrifugal force at the top is greater than or equal to the force exerted due to the weight of the Yo-Yo (mass of string is considered negligible here). Only then there is some tension in the string to prevent it from slackening.



          beginalignrequirecancelcancelm,fracv^2r &ge cancelmg\ fracv^2r &ge g\ v &gesqrtgrendalign



          $m$: mass of the Yo-Yo, $v$: velocity at highest point,



          $r$: vertical radius of the circle (or length of the string) and $g$ is acceleration due to gravity.



          Since you are looking for minimum speed, we can take minimum value of $v$ , that is



          $sqrtgr=sqrt(9.8times1.5)=3.83largefracmathrmmmathrmsec$



          which is basically independent of the mass of the yo-yo.






          share|cite|improve this answer






















          • @santimirandarp is not not possible to get 'greater than or equal to' symbol up there in the first three equation?
            – user64829
            2 hours ago










          • Is it better now? You can edit it by yourself by analogy with the symbols I've used!
            – santimirandarp
            1 hour ago










          • Yeah, much better. Sorry for the trouble, I'm writing here for the first time.
            – user64829
            47 mins ago












          up vote
          3
          down vote










          up vote
          3
          down vote









          If the Yo-Yo string doesn't slacken, the centrifugal force at the top is greater than or equal to the force exerted due to the weight of the Yo-Yo (mass of string is considered negligible here). Only then there is some tension in the string to prevent it from slackening.



          beginalignrequirecancelcancelm,fracv^2r &ge cancelmg\ fracv^2r &ge g\ v &gesqrtgrendalign



          $m$: mass of the Yo-Yo, $v$: velocity at highest point,



          $r$: vertical radius of the circle (or length of the string) and $g$ is acceleration due to gravity.



          Since you are looking for minimum speed, we can take minimum value of $v$ , that is



          $sqrtgr=sqrt(9.8times1.5)=3.83largefracmathrmmmathrmsec$



          which is basically independent of the mass of the yo-yo.






          share|cite|improve this answer














          If the Yo-Yo string doesn't slacken, the centrifugal force at the top is greater than or equal to the force exerted due to the weight of the Yo-Yo (mass of string is considered negligible here). Only then there is some tension in the string to prevent it from slackening.



          beginalignrequirecancelcancelm,fracv^2r &ge cancelmg\ fracv^2r &ge g\ v &gesqrtgrendalign



          $m$: mass of the Yo-Yo, $v$: velocity at highest point,



          $r$: vertical radius of the circle (or length of the string) and $g$ is acceleration due to gravity.



          Since you are looking for minimum speed, we can take minimum value of $v$ , that is



          $sqrtgr=sqrt(9.8times1.5)=3.83largefracmathrmmmathrmsec$



          which is basically independent of the mass of the yo-yo.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 47 mins ago









          santimirandarp

          342112




          342112










          answered 3 hours ago









          user64829

          518




          518











          • @santimirandarp is not not possible to get 'greater than or equal to' symbol up there in the first three equation?
            – user64829
            2 hours ago










          • Is it better now? You can edit it by yourself by analogy with the symbols I've used!
            – santimirandarp
            1 hour ago










          • Yeah, much better. Sorry for the trouble, I'm writing here for the first time.
            – user64829
            47 mins ago
















          • @santimirandarp is not not possible to get 'greater than or equal to' symbol up there in the first three equation?
            – user64829
            2 hours ago










          • Is it better now? You can edit it by yourself by analogy with the symbols I've used!
            – santimirandarp
            1 hour ago










          • Yeah, much better. Sorry for the trouble, I'm writing here for the first time.
            – user64829
            47 mins ago















          @santimirandarp is not not possible to get 'greater than or equal to' symbol up there in the first three equation?
          – user64829
          2 hours ago




          @santimirandarp is not not possible to get 'greater than or equal to' symbol up there in the first three equation?
          – user64829
          2 hours ago












          Is it better now? You can edit it by yourself by analogy with the symbols I've used!
          – santimirandarp
          1 hour ago




          Is it better now? You can edit it by yourself by analogy with the symbols I've used!
          – santimirandarp
          1 hour ago












          Yeah, much better. Sorry for the trouble, I'm writing here for the first time.
          – user64829
          47 mins ago




          Yeah, much better. Sorry for the trouble, I'm writing here for the first time.
          – user64829
          47 mins ago

















           

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