Conservation of energy - velocity problem

• lolzwhut?
In summary: That's the easy way to do it. The harder way would be to calculate the velocity of the particle at the bottom with the equation v = sqrt(2 * 9.8 * 26.7). Then plug that into the kinematics equation: v(final)^2 = v(initial)^2 + 2ax, where v(initial) is what you calculated in the previous equation. Solve for...v(final) = sqrt(2 * 9.81 * 26.7) = sqrt(2 * 9.81) + 2a v(initial) = sqrt(2 * 9.8) + 2a
lolzwhut?

Homework Statement

A bead slides without friction around a loop-
the-loop. The bead is released from a height
of 26.7 m from the bottom of the loop-the-
loop which has a radius 9 m.
The acceleration of gravity is 9.8 m/s2 .

Diagram:
http://img31.imageshack.us/img31/2377/19550530.png

What is its speed at point A?
Answer in units of m/s.

a of grav = 9.8 m/^2
released from: 26.7 m
loop radius 9 m

PE = mgh
Ke = .5mv^2

The Attempt at a Solution

Ok so this is what I did. I set the PE and KE equations equal to each other so mass can cancel out because its not given.

MGH = .5mv2 >>>> which turns out to be >>> V=sqrt(2*g*h)
To get height i did 26.7-9= 17.7

Then I plugged it into the equation:
v = sqrt(2 * 9.81 * 17.7)
= 18.64 m/s

I submitted the answer and it was wrong still!

omg, someone please explain what am i doing wrong! physics test soon :(

Last edited by a moderator:
What is the total energy of the system?

At point A, what types of energy does the bead have?

rock.freak667 said:
What is the total energy of the system?

At point A, what types of energy does the bead have?

Wouldn't KE be 0? I'm not sure how to do this one..I'm really confused

lolzwhut? said:
Wouldn't KE be 0? I'm not sure how to do this one..I'm really confused

nope. KE is not zero, so what is an expression for the energy at point A?

rock.freak667 said:
nope. KE is not zero, so what is an expression for the energy at point A?

mgh=1/2mv^2?

Velocity is 0 at that point becuse its the maximum

lolzwhut? said:
mgh=1/2mv^2?

Velocity is 0 at that point becuse its the maximum

No. The velocity is not zero. At point A, there is are two types of energy present. The initial energy is converted into these two types.

what about the rotational kinetic energy?

rock.freak667 said:
No. The velocity is not zero. At point A, there is are two types of energy present. The initial energy is converted into these two types.

Potential Energy and Kinetic Energy yes?Thats why I set them equal to to cancel out mass. But its still incorrect.

i think if the object is ROTATING ( because i haven't understood if it is sliding means rotating or not) then
PE= linear KE+ Rotational KE.
maybe this can help us!
i am not sure anyway..we had just learned about it.

lolzwhut? said:
Potential Energy and Kinetic Energy yes?Thats why I set them equal to to cancel out mass. But its still incorrect.

Good, at A there is mgh+ 1/2 mv2 put this equal to the initial energy now.

rock.freak667 said:
Good, at A there is mgh+ 1/2 mv2 put this equal to the initial energy now.

MGH = .5mv2 >>>> which turns out to be >>> V=sqrt(2*g*h)

wouldn't that be correct?

lolzwhut? said:
MGH = .5mv2 >>>> which turns out to be >>> V=sqrt(2*g*h)

wouldn't that be correct?

you'd have

mgH=mgh+1/2mv2

rock.freak667 said:
you'd have

mgH=mgh+1/2mv2

But we'd still have 1 mass left in the equation. Or do all 3 of them cancel out x_x

lolzwhut? said:
But we'd still have 1 mass left in the equation. Or do all 3 of them cancel out x_x

one bead...all the same mass

rock.freak667 said:
one bead...all the same mass

I don't understand what you mean lol. It doesn't give us a mass. So why use mass in the equation when not even all of them cancel out. How am I supposed to solve it with that equation. I'm confused xD

lolzwhut? said:

Homework Statement

A bead slides without friction around a loop-
the-loop. The bead is released from a height
of 26.7 m from the bottom of the loop-the-
loop which has a radius 9 m.
The acceleration of gravity is 9.8 m/s2 .

Diagram:
http://img31.imageshack.us/img31/2377/19550530.png

What is its speed at point A?
Answer in units of m/s.

a of grav = 9.8 m/^2
released from: 26.7 m
loop radius 9 m

PE = mgh
Ke = .5mv^2

The Attempt at a Solution

Ok so this is what I did. I set the PE and KE equations equal to each other so mass can cancel out because its not given.

MGH = .5mv2 >>>> which turns out to be >>> V=sqrt(2*g*h)
To get height i did 26.7-9= 17.7

Then I plugged it into the equation:
v = sqrt(2 * 9.81 * 17.7)
= 18.64 m/s

I submitted the answer and it was wrong still!

omg, someone please explain what am i doing wrong! physics test soon :(

This discussion is getting too cluttered when you're already on the right path. The only mistake you made here is that you got the height wrong. It should be 26.7 - 18 = 8.7 meters. Remember, you want the height difference between your original point and your final point (so if radius is 9, the diameter of the hoop is 18). You should get 13.1 m/s.

That's the easy way to do it. The harder way would be to calculate the velocity of the particle at the bottom with the equation v = sqrt(2 * 9.8 * 26.7). Then plug that into the kinematics equation: v(final)^2 = v(initial)^2 + 2ax, where v(initial) is what you calculated in the previous equation. Solve for v(final) and that would also give you the same answer.

Last edited by a moderator:
Cryxic said:
This discussion is getting too cluttered when you're already on the right path. The only mistake you made here is that you got the height wrong. It should be 26.7 - 18 = 8.7 meters. Remember, you want the height difference between your original point and your final point (so if radius is 9, the diameter of the hoop is 18). You should get 13.1 m/s.

, thanks a ton man. I think this was a trick question...Since I never thought of the diameter. It makes sense why its 26.7-18 instead of 9.

Thanks for clearing it up, I was really starting to get a little confused there.

1. What is the conservation of energy?

The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but only transferred or converted from one form to another.

2. How is velocity related to the conservation of energy?

In the context of the conservation of energy, velocity refers to the speed at which an object is moving in a certain direction. The conservation of energy principle states that the total energy of a system, including kinetic energy (which is related to velocity), remains constant over time.

3. Can the conservation of energy be violated?

No, the conservation of energy is a fundamental law of physics and has been extensively tested and proven to hold true in all known physical systems. Any apparent violations of this principle can be explained by factors such as external forces or energy losses to heat or friction.

4. How is the conservation of energy applied in real-world situations?

The conservation of energy is applied in various fields such as engineering, mechanics, and thermodynamics. It is used to analyze and solve problems involving energy transfer and transformation in systems such as machines, vehicles, and power plants.

5. What are some common misconceptions about the conservation of energy?

One common misconception is that the conservation of energy only applies to closed systems, when in fact it applies to all physical systems. Another misconception is that energy is "used up" or "lost" in a system, when in reality it is simply converted from one form to another.

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