Conservation of energy - velocity problem

[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

Homework Equations

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 :(

 

[rock.freak667]

What is the total energy of the system?

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

 

[lolzwhut?]

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

 

[rock.freak667]

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?

 

[lolzwhut?]

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

 

[rock.freak667]

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.

 

[mikhailpavel]

what about the rotational kinetic energy?

 

[lolzwhut?]

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.

 

[mikhailpavel]

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.

 

[rock.freak667]

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.

 

[lolzwhut?]

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?

 

[rock.freak667]

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

wouldn't that be correct?

you'd have

mgH=mgh+1/2mv²

 

[lolzwhut?]

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

 

[rock.freak667]

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

 

[lolzwhut?]

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

 

[Cryxic]

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

Homework Equations

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.

 

[lolzwhut?]

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.