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5.2.2Torque and Rotational Dynamics

Force Diagrams and Torque — AP PHYSICS 1 Revision Notes & Topical Questions

5 min read 9 parts Lesson 11 of 18 1 quiz

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Lesson summary

5.2.2 Force Diagrams and Torque — AP PHYSICS 1 notes: Draw force diagrams for extended objects, locate each force's point of application and use them to find net torque and equilibrium conditions.Force diagram — like a free-body diagram, with arrow length for relative magnitude a…

  • Difference
  • Pulley system
  • Contact forces
  • Hinged rod example
  • Pivot trick
  • Translational condition

Part 1 of 7Force diagrams

Force diagrams
Force diagram
  • Force diagram like a free-body diagram, with arrow length for relative magnitude and arrow direction for direction.
Difference
  • Difference free-body diagrams treat the object as a point particle;
    • force diagrams treat it as a rigid system.
Purpose
  • Purpose they show where forces act relative to the axis, allowing prediction of rotation.
Pulley system
  • Pulley system use free-body diagrams for the hanging masses and a force diagram for the rotating pulley.

Part 2 of 7Drawing a force diagram of an extended object

Object
  • Object draw the beam, door or ladder as a line or box.
Arrows
  • Arrows each force starts at its point of application.
Weight
  • Weight at the centre of mass.
    • (geometric centre if uniform)
Contact forces
  • Contact forces normal forces at contact points, tension at attachments, friction along surfaces, applied forces, hinge force.
Hinged rod example
  • Hinged rod example forces are the hinge force, the cord tension at an angle, the rod's weight at its centre and the block's weight at the end.

Part 3 of 7Computing net torque

1
Draw the object and mark the chosen axis (pivot).
2
Draw each force at its point of application with magnitude and direction.
3
Find r and the angle θ between r and F, or directly the lever arm r⊥.
4
Find each torque with τ = rF sinθ, positive if counterclockwise, negative if clockwise.
5
Add: Στ = Στccw − Στcw.
  • Pivot trick
    • a force whose line of action passes through the axis gives zero torque, so choosing the pivot at the hinge removes the hinge force.

Part 4 of 7Static equilibrium

Translational condition
  • ΣFx = 0 and ΣFy = 0.
Rotational condition
  • Στ = 0 about any axis, so choose the simplest axis, usually where unknown forces act.
See-saw
  • See-saw clockwise torques equal counterclockwise torques, m1gr1 = m2gr2.

Part 5 of 7Weight and the centre of mass

1Uniform beam of length L pivoted at one end
  • Uniform beam of length L pivoted at one end weight torque is mg(L/2).
2Ladder against a wall
  • Ladder against a wall weight acts at the midpoint.
3Hanging mass
  • Hanging mass its weight acts at the point of attachment.
4Nonuniform object
  • Nonuniform object centre of mass lies closer to the heavy end and may need to be found.

Part 6 of 7Net torque and rotation

Unbalanced torques
  • Unbalanced torques the object rotates with angular acceleration in the direction of the net torque.
Cut string
  • Cut string a horizontal beam hinged at one end rotates downward due to the clockwise torque of its weight.
Size of α
  • Size of α set by Στ = Iα.
Limiting cases
  • Limiting cases force at the pivot gives zero torque;
    • a perpendicular force at the far end gives the largest torque for that force.
Tabulating
  • Tabulating list force, lever arm, sign and torque, then sum.

Part 7 of 7More from the specification

More from the specification
Distance
  • Total length of path travelled, a scalar.
Vector
  • Quantity with magnitude and direction.
Component
  • Part of a vector along one axis.
Perpendicular distance
  • Distance from the axis to the line of motion, r⊥ = r sinθ.

Key terms

TermMeaning
Static equilibriumNet force and net torque both zero with the object at rest.
Centre of massPoint where the weight can be taken to act.
PivotAxis about which torques are calculated.
Line of actionLine along which a force acts.
Hinge forceForce a hinge exerts on a beam.

Avoid these mistakes

1
Placing the weight at the end of the object rather than at the centre of mass.
2
Calculating torques about different axes in the same equation.
3
Forgetting the hinge reaction force, or that it gives no torque about the hinge.
4
Requiring only ΣF = 0 for equilibrium without Στ = 0.
5
Using the length of the object rather than the distance to the point of application as r.

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