Solving for V in mgh=1/2mv^2+1/2mr^2(v/r)^2

  • Thread starter Jordash
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In summary, the conversation is about solving an equation for the variable v using algebra. The equation is mgh=1/2mv^2+1/2mr^2(v/r)^2 and the person is looking for assistance in understanding the steps to solve for v. A hint is given and a website is recommended for learning algebra. The conversation also touches on finding resources for learning Physics, with the exception of one website that is highly recommended.
  • #1
Jordash
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Homework Statement



mgh=1/2mv^2+1/2mr^2(v/r)^2

I need to solve for v, I'm really stuck on Algebra can anyone show me the steps I'd take to solve for V?

Thanks

Homework Equations





The Attempt at a Solution



Tried thinking it through but I'm getting stuck :( :(
 
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  • #2
Here's a hint:

[tex](\frac{v}{r})^2 = \frac{v^2}{r^2}[/tex]

also, if the variable m on the left hand side of the equation is the exact same m as on the right side, you can divide both sides by m to get rid of it.
 
  • #3
Thanks for the help i'll try it, anyone know of resources that help you learn Algebra like this, it's really confusing to me.

Thanks
 
  • #4
Try this site: http://www.wtv-zone.com/Angelaruth49/Algebra.html

teaches some basic algebra.
 
  • #5
Cool thank you, And does anyone know a good site to learn Physics it's all going over my head in class.
 
  • #6
google.
 
  • #7
Sorry to bash on Physics websites but so far every Physics site I've found has had terrible design, poor teaching and organization. (similar to my teaching) With the exception of this site. (Best Physics Site I've found). I've found great resources for learning on every other subject through Google except for Physics.
 

1. How do I solve for V in the equation mgh=1/2mv^2+1/2mr^2(v/r)^2?

To solve for V in this equation, you will need to use algebraic manipulations to isolate the variable V on one side of the equation. Begin by distributing the 1/2mr^2 term on the right side of the equation. Then, factor out the common term of V^2 and use the inverse operations to isolate V. Finally, use the square root function to find the value of V.

2. What do the variables in this equation represent?

The variable m represents the mass of the object, g represents the acceleration due to gravity, h represents the height from which the object is dropped, and r represents the radius of the circular path the object travels on. V represents the final velocity of the object.

3. Why is this equation important in science?

This equation is important in science because it relates the potential energy of an object (mgh) to its kinetic energy (1/2mv^2). It also takes into account the rotational kinetic energy of the object (1/2mr^2(v/r)^2) if it is moving in a circular path. This equation is often used in physics and engineering to analyze the motion and energy of objects.

4. Can this equation be used to solve for V in any situation?

Yes, this equation can be used to solve for V in any situation where the object is moving under the influence of gravity and/or rotating in a circular path. However, it is important to note that this equation assumes certain conditions such as negligible air resistance and a constant gravitational field.

5. Are there any other equations that are related to this one?

Yes, there are other equations that are related to this one. For example, the equation for gravitational potential energy (PE=mgh) is a component of this equation. Additionally, the equation for centripetal acceleration (a=v^2/r) is used to find the value of v/r in the rotational kinetic energy term (1/2mr^2(v/r)^2) of this equation.

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