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.

homework-and-exercises newtonian-mechanics

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edited 1 hour ago

Qmechanic♦

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asked 3 hours ago

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up vote
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1

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

edited 1 hour ago

Qmechanic♦

97.9k121661059

asked 3 hours ago

Archipelago2000

1032

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

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

edited 1 hour ago

Qmechanic♦

97.9k121661059

asked 3 hours ago

Archipelago2000

1032

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

share|cite|improve this question

edited 1 hour ago

Qmechanic♦

97.9k121661059

asked 3 hours ago

Archipelago2000

1032

share|cite|improve this question

edited 1 hour ago

Qmechanic♦

97.9k121661059

asked 3 hours ago

Archipelago2000

1032

share|cite|improve this question

share|cite|improve this question

edited 1 hour ago

Qmechanic♦

97.9k121661059

edited 1 hour ago

Qmechanic♦

97.9k121661059

edited 1 hour ago

Qmechanic♦

97.9k121661059

97.9k121661059

asked 3 hours ago

Archipelago2000

1032

asked 3 hours ago

Archipelago2000

1032

asked 3 hours ago

Archipelago2000

1032

1032

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

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

edited 40 mins ago

santimirandarp

342112

answered 2 hours ago

user64829

518

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

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

edited 40 mins ago

santimirandarp

342112

answered 2 hours ago

user64829

518

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

edited 40 mins ago

santimirandarp

342112

answered 2 hours ago

user64829

518

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

edited 40 mins ago

santimirandarp

342112

answered 2 hours ago

user64829

518

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

edited 40 mins ago

santimirandarp

342112

answered 2 hours ago

user64829

518

share|cite|improve this answer

share|cite|improve this answer

edited 40 mins ago

santimirandarp

342112

edited 40 mins ago

santimirandarp

342112

edited 40 mins ago

santimirandarp

342112

342112

answered 2 hours ago

user64829

518

answered 2 hours ago

user64829

518

answered 2 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
  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
  39 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

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

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

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
39 mins ago

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

add a comment | 

 

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