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

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A 100g block attached to a 50cm wire is analyzed for its velocity and tension when it reaches equilibrium and at a 30° angle with the vertical. Using conservation of mechanical energy, the block's velocity at equilibrium is calculated to be 3.13 m/s. The tension in the wire is determined by considering the forces acting on the block, leading to the conclusion that tension equals the centripetal force minus the gravitational force. The correct tension at the equilibrium position is found to be 0.98 N after accounting for these forces. The discussion emphasizes the importance of accurately applying Newton's second law to derive the tension in the wire.
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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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