This is way easy if you are smart

  • Thread starter emerica86
  • Start date
In summary, the minimum speed required for the rock to just make it around the top of the circle is 9.81 [m/s^2]. The forces acting on the rock at the top of the circle are gravity and the centripetal force, which is equal to the weight of the rock. The equation for the centripetal force is f=mv^2/r, and since weight is equal to mg, the mass of the rock cancels out and we are left with the equation mg=mv^2/r. By solving for v, we can find the minimum speed needed. It is helpful to start with a free body diagram and sum the forces in the x and y directions.
  • #1
emerica86
10
0
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



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
 
Last edited:
Physics news on Phys.org
  • #2
Okay, at the top of the circle what forces are acting on the rock?
 
  • #3
gravity ...?
 
  • #4
emerica86 said:
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?
 
  • #5
gravity i think but i tried 9.8m/s and it was wrong...
 
  • #6
emerica86 said:
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.
 
  • #7
Any object undergoing circular motion must experience a C########## Force.
 
  • #8
i don't know what forces are acting on it i am so confused
 
  • #9
oh a centriipugal force?
 
  • #10
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.
 
  • #11
emerica86 said:
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?
 
  • #12
f=mv^2/r ?
 
  • #13
emerica86 said:
f=mv^2/r ?
That's the one :smile:, 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?
 
  • #14
yes i follow i just don't get what the weight is because the problem says nothign to do with weight.
 
  • #15
emerica86 said:
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 ([itex]W=m\cdot g[/itex])
 
  • #16
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?
 
  • #17
emerica86 said:
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...
 
  • #18
omg i give up unless someone just gives me the answer lol
 
  • #19
emerica86 said:
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;

[tex]mg = \frac{mv^2}{r}[/tex]

Can you go from here?
 
  • #20
oh duhh
ok thank you very much
 
  • #21
emerica86 said:
oh duhh
ok thank you very much
My pleasure :smile:
 
  • #22
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.
 

Related to This is way easy if you are smart

1. What does "This is way easy if you are smart" mean?

This phrase implies that the task at hand is simple or uncomplicated, but may require a certain level of intelligence or skill to complete successfully.

2. Is this phrase offensive or derogatory towards those who may struggle with the task?

It can be seen as offensive or insensitive to those who may not find the task easy, as it implies that they are not smart enough to complete it. It is important to use inclusive language and avoid making assumptions about someone's intelligence.

3. Can someone who is not naturally smart still find the task easy?

Yes, intelligence is not solely determined by natural ability. Someone who may not have a high IQ can still excel at the task by putting in effort and utilizing effective strategies.

4. What impact does labeling a task as "easy if you are smart" have on individuals who may not find it easy?

This label can create a sense of discouragement or demotivation for those who may struggle with the task. It can also perpetuate the harmful stereotype that intelligence is fixed and cannot be improved upon.

5. How can we rephrase this phrase to be more inclusive and respectful?

A better way to phrase this would be to focus on the task itself and acknowledge that it may require certain skills or knowledge to complete. For example, "This task may be challenging, but it can be successfully completed with the right strategies and effort."

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
3K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
19
Views
895
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
12K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
8K
Back
Top