Resource Lesson
Tension Cases: Four Special Situations
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Case I: Static Equilibrium
- There is no acceleration in either the x- or y-dimension. In each case, ΣF = 0.
ΣF
_{x}
= 0 → T
_{cord}
+ (-T sin θ) = 0
ΣF
_{y}
= 0 → T cos θ - mg = 0
Refer to the following information for the next two questions.
Let the block has a mass of 5 kg and θ equal 37º in the above diagram.
What is the tension in the diagonal rope?
What is the tension in the cord?
Case II: Conical Pendulum
- The forces are balanced in the y-dimension but there is an unbalanced x-dimension force directed towards the center of the circle.
ΣF
_{x}
= ma
_{c}
→ T sin θ = ma
_{c}
ΣF
_{y}
= 0 → T cos θ - mg = 0
Refer to the following information for the next two questions.
Let the block has a mass of 5 kg, θ equal 37º, and the length of the string equal 1.2 meters in the above diagram.
What is the tension in the string?
How fast is the block moving along its circular path?
Hint:
r = L sin θ
Case III: Vertical Circles
- In vertical circular motion, the acceleration is not uniform as gravity speeds up objects while they fall and slows them down as they rise. Tension is greatest at the bottom of a vertical circle and approach minimum values while passing through the top of a vertical circle.
top
bottom
top
bottom
net F
_{c}
= T + mg
net F
_{c}
= T - mg
m(v
^{2}
/r) = T + mg
m(v
^{2}
/r) = T - mg
T = m(v
^{2}
/r) - mg
T = m(v
^{2}
/r) + mg
The following formula used to calculate the
minimum, or critical, speed
required for the block to pass through the top of a vertical circle is derived by taking the limit as T → 0 in the previous formula for centripetal force at the top of a vertical circle and solving for v:
m(v
^{2}
/r) = mg
v
^{2}
/r = g
v
^{2}
= rg
v
_{critical}
= √(rg)
Refer to the following information for the next two questions.
Let the block has a mass of 5 kg and the radius equal 1.2 meters in the above diagrams.
What is the minimum speed needed for the block to maintain its circular path it passes through the top of the circle?
Using the answer to the previous question, what would the tension be in the string as the block passes through the bottom of the circle?
Hint:
Use conservation of energy to calculate the speed of the block at the bottom of the circle.
Case IV: Simple Pendulum
- As a simple pendulum swings, its bob experiences a tangential acceleration along its arc and a centripetal acceleration towards the center of the circle. Remember that a pendulum is merely the bottom of a vertical circle. Energy methods should be used to calculate the velocity at ang given time. To review this procedure, visit this related lesson on
energy conservation in simple pendulums
.
ΣF
_{x}
= ma
_{t}
→ mg sin θ = ma
_{t}
ΣF
_{y}
= ma
_{c}
→ T - mg cos θ = ma
_{c}
Refer to the following information for the next six questions.
In the diagram shown above, assume that a 5-kg mass is released from rest at θ equal to 37º (measured from equilbrium).
At the exact instant of release would the mass experience any centripetal acceleration?
At the exact instant of release what is the instantaneous tension in the string?
At the exact instant of release what is the instantaneous tangential acceleration pulling the mass towards equilibrium when θ = 37º?
Why does the term "instantaneous" have to be inserted in these questions?
As the pendulum loses potential energy and gains kinetic energy (prior to its arrival at equilibrium), does the tension in the string increase, decrease, or remain constant?
As the pendulum swings towards equilibrium, the bob is being accelerated both along the tangent and centripetally towards the center. Sketch a diagram to show the direction of its resultant acceleration.
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