This is way easy if you are smart

[emerica86]

circle question its easy (still need help)

Suppose you tie a rock to the end of a 1.5 m long string and spin it in a vertical circle. What is the minimum speed it can travel and "just" get around the top? (Hint: The rope will just start to go "slack" and the only force acting on the rock, downward, is gravity.)

-------m/s

:yuck:

here is my work:
so i thought it would be 9.9 since 9.8 is gravity or something i am completely lost. if someone could help that would be great

 

[Hootenanny]

Okay, at the top of the circle what forces are acting on the rock?

 

[emerica86]

gravity ...?

 

[Hootenanny]

gravity ...?

Yes (assuming the string is slack as the hint suggests), so since the rock is traveling in a circle what must the weight of the rock be equal to?

 

[emerica86]

gravity i think but i tried 9.8m/s and it was wrong...

 

[radou]

gravity i think but i tried 9.8m/s and it was wrong...

9.81 [m/s^2] is gravitational acceleration, and Hootenanny asked you about a force.

 

[Hootenanny]

Any object undergoing circular motion must experience a C########## Force.

 

[emerica86]

i don't know what forces are acting on it i am so confused

 

[emerica86]

oh a centriipugal force?

 

[JSBeckton]

Draw a free body diagram, 2 forces, centrifigal force and gravity, they are equal at the critical point. f=ma for gravity, i can't remember the centrifigal force eqn off the top of my head but mass should cancel and you can solve for velocity.

Its been awhile since intro physics so I'm a bit rusty.

 

[Hootenanny]

oh a centriipugal force?

No, centrifugal forces only arise when we consider the circular motion from an non-inertial reference frame (i.e. the rocks), the correct term would be centripetal force. Now, what equations do you know that deal with the centripetal force?

 

[emerica86]

f=mv^2/r ?

 

[Hootenanny]

f=mv^2/r ?

That's the one , so since weight is the only force acting when the rock is at the minimum velocity, it follows that the weight of the rock must be providing all the required centripetal force at this point. Do you follow?

 

[emerica86]

yes i follow i just don't get what the weight is because the problem says nothign to do with weight.

 

[Hootenanny]

yes i follow i just don't get what the weight is because the problem says nothign to do with weight.

I'm sure you know that the weight of an object is the product of it's mass and the gravitational field strength ($W=m\cdot g$)

 

[emerica86]

yes i know that equation I am just saying the origional problem dosent tell you the mass of the rock. so how do you get it?

 

[Hootenanny]

yes i know that equation I am just saying the origional problem dosent tell you the mass of the rock. so how do you get it?

The question is do you really need it? Equate the weight of the rock with your equation for the centripetal force and see what happens to the mass...

 

[emerica86]

omg i give up unless someone just gives me the answer lol

 

[Hootenanny]

omg i give up unless someone just gives me the answer lol

Okay, I'll give you a hint (but don't be too quick to abandon hope). If we equate the weight of the rock with the centripetal force we obtain;

$$mg = \frac{mv^2}{r}$$

Can you go from here?

 

[emerica86]

oh duhh
ok thank you very much

 

[Hootenanny]

oh duhh
ok thank you very much

My pleasure

 

[JSBeckton]

Always start with a FBD and just look at the situation, often times the picture will make it easier to understand. Sum the forcs in the x and y directions, this will tell you what the forces must be, then all you have to do is find the rigth formulas and solve.