Energy of a system-a box moving in a circle

haha1234
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Homework Statement



A light spring has unstressed length 15.5cm.It is described by Hooke's law with spring constant 4.30N/m.One end of the horizontal spring is held on a fixed vertical axle,and the other end is attached to a punk of mass that can move without friction over a horizontal surface.The punk is set into motion in a circle with a period of 1.30s.
Find the extension of the spring x as it depends on m.

Homework Equations





The Attempt at a Solution



Firstly,I found the velocity of the punk.
1/2kx2=1/2mv2
v=[(4.3x2)/m]1/2
But I don't know what is the next step.:cry:
 
on Phys.org
Let r be the radius of the circle. What is the radial acceleration of the mass? If the mass is traveling at this radius, how much is the spring stretched? What is the centripetal force that the spring must supply?
 
I don't think you are using the right equation here. Your equation is saying that the energy in the spring equals the kinetic energy of the punk - why would that be the case? Do you know any other equations that might apply?
 
Jilang said:
I don't think you are using the right equation here. Your equation is saying that the energy in the spring equals the kinetic energy of the punk - why would that be the case? Do you know any other equations that might apply?

I have used this approach is because the question has given the period of the circle,so I think I should find the velocity...
 
OK, do you know any other equations that have velocity in them?
 
What's the angular velocity ω?
 
chestermiller said:
what's the angular velocity ω?

ω=360°/1.3
=277?
andω=4.3/m?
 
Last edited:
haha1234 said:
ω=360°/1.3
=277?
andω=4.3/m?
The angle has to be in radians, not degrees. After you get the angular velocity correct, express the radial acceleration in terms of ω and r.

Chet
 
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Chestermiller said:
The angle has to be in radians, not degrees. After you get the angular velocity correct, express the radial acceleration in terms of ω and r.

Chet

I think I cannot get the correct radical acceleration.
The radical acceleration I have got is a=(4∏2/1.69)x
 
  • #10
haha1234 said:
I think I cannot get the correct radical acceleration.
The radical acceleration I have got is a=(4∏2/1.69)x

Correct...It would be better if you express it as a=(4∏2/1.69)(.155+x) where x is the extension in the spring .
 
  • #11
Tanya Sharma said:
Correct...It would be better if you express it as a=(4∏2/1.69)(.155+x) where x is the extension in the spring .

Why the radius of the circle is 0.155+x?
 
  • #12
haha1234 said:
Why the radius of the circle is 0.155+x?

What do you think radius would be ?
 
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