Calculate Velocity & Tension of 100g Block on 50cm Wire

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Homework Help Overview

The problem involves a 100g block attached to a 50cm wire, initially at rest, and requires calculating the block's velocity and the tension in the wire at equilibrium and at a 30° angle with the vertical. The context is centered around concepts of mechanical energy conservation and forces acting on the block.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using conservation of mechanical energy to find the block's velocity at equilibrium and question how to calculate the tension in the wire. Some suggest considering centripetal acceleration and the forces acting on the block.

Discussion Status

There is ongoing exploration of the relationships between tension, gravitational force, and centripetal acceleration. Participants are clarifying their understanding of the forces involved and how to apply Newton's second law in this context. Multiple interpretations of the tension calculation are being discussed.

Contextual Notes

Participants note the importance of drawing diagrams to visualize forces and clarify assumptions about the equilibrium position and the direction of acceleration.

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Homework Statement


A block of 100g is attached to a wire inexstensible long 50cm. The block is left with a null initial velocity when the wire form an angle of 90° with the vertical. Calculate the velocity of the block and the tension of the wire when the block is at the position of equilibrium. Also calculate the velocity of the block when it forms an angle of 30° with the vertical after the position of equilibrium.

Homework Equations

The Attempt at a Solution


I started by calculating the velocity of the block at the position of equilibrium.
I used the law of conservation of mechanical energy and so:
K0 = 1/2 * m * v02 = 0
because the initial velocity is zero.
U0 = 0
K = 1/2 * m * v2
U = - m * g * l
So we have
0 =1/2 * m * v2 - m * g * l
v = sqrt(2 * g * l) = 3.13 m/s
After this I should calculate the tension of the wire too but I don't get how to do it.
I tried doing this:
{m * ax = - T sin90° = - T
{m * ay = T cos90° - P = - P = m * g
But beside this I don't know how to proceed.
Can someone help me how to do this little part?
 
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Maybe by the sum of the forces acting on the block at the bottom needs to provide its centripetal acceleration?
 
andrevdh said:
Maybe by the sum of the forces acting on the block at the bottom needs to provide its centripetal acceleration?
So you are telling me that the tension of the wire is T = m * ac?
Because these way it would be ac = v2/l = 19.6 m/s2, so T = 0.1kg * 19.6 m/s2 = 1.96 N
Is it correct?
 
The weight is also "contributing"
 
So it is T = m * ac - m * g = 0.98 N?
 
Draw a little diagram with the forces and maybe take positive as up (the direction of the acceleration)
 
andrevdh said:
Draw a little diagram with the forces and maybe take positive as up (the direction of the acceleration)
I did it and the acceleration is positive so the counts are right, aren't they? I don't get why you asked me that.
 

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The equilibrium position is at the bottom. I assumed you wanted to calculate the tension in the wire at this point in the swing.
 
andrevdh said:
The equilibrium position is at the bottom. I assumed you wanted to calculate the tension in the wire at this point in the swing.
Yeah, that's the initial position. The right one is this:
 

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  • #10
Yes, so if you apply N2 and take up as positive then T - P = mac
 
  • #11
Oh! It's m * ac = T - P, not T = m * ac - P. I thought that the tension was directly the centripetal force minus the gravitational force and not that the centripetal force was the difference of the other two.
Thank you for your help!
 

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