A body with that is not accelerating is in equilibrium. A body travelling at constant speed is in translational equilibrium. A body rotating at constant angular velocity is in rotational equilibrium.
Key Concepts
A body is in rotational equilibrium if the sum of the anticlockwise torques is equal to the sum of the clockwise torques about a pivot:
\(\sum\Gamma_\textrm{anticlockwise}= \sum\Gamma_\textrm{clockwise}\)
A body is in translational equilibrium if no resultant force acts:
\(\sum F_\textrm{upward}=\sum F_\textrm{downward}\)
\(\sum F_\textrm{left}=\sum F_\textrm{right}\)
We can use these rules in combination to solve problems about objects on a pivot.
In this example, body is in rotational and translational equilibrium if the following apply:
- \(Aa=Bb\) (using torques)
- \(P=A+B\) (considering vertical forces)
Note that the beam has no mass. If the beam has mass, it the force \(mg\) acts vertically downwards in the centre.
Here's an example of a rod in equilibrium.
Placed on two supports, our rod becomes a bridge.
Like torque for force and moment of inertia for mass, there is also a rotational equivalent for acceleration, angular acceleration.
Angular acceleration is defined as the rate of change of angular velocity:
\(a={\Delta \omega\over \Delta t}={{\omega _f-\omega_i}\over t}\)
- \(\alpha\) is angular acceleration (rads-2)
- \(\omega_f\) is final angular velocity (rads-1)
- \(\omega_i\) is initial angular velocity (rads-1)
- \(t\) is the time taken (s)
We can convert between tangential and angular acceleration:
\(a=r\alpha\)
- \(a\) is tangential acceleration (ms-2)
- \(r\) is the radius of the circle (m)
- \(\alpha\) is angular acceleration (rads-2)
Recall from Oscillations and Circular motion, there are two methods for finding angular velocity:
- Angular velocity is the instantaneous rate of change of angular displacement (rad): \(\omega={\Delta \theta \over \Delta t}\)
- For repeated circular motion, angular velocity is the total angle moved through for a complete circle (\(2\pi\) rad) divided by the time period (s): \(\omega = {2\pi \over T}=2\pi f\), where \(f\) is the frequency (Hz).
Note the three forms of acceleration:
- Angular acceleration is the rate of change of angular velocity, or the rate of change in the angle subtended.
- Angular acceleration is is proportional to the tangential acceleration, the rate of change of translational velocity. This requires a resultant force acting along a tangent to the circle.
- These are not related to centripetal acceleration. Centripetal acceleration is towards the centre of the circle and thus requires a resultant force along a radius to the circle, perpendicular to the tangent.
Use flashcards to practise your recall.
Use quizzes to practise application of theory.
START QUIZ!
A ladder leans against a wall.
Which of the following forces are equal?
The ladder is in equilibrium so vertical forces are balanced.
NOTE: \(F\) and \(N_t\) are also equal.
A ladder leans against a wall.
The torque caused by the weight about the top of the ladder is
Torque = force x perpendicular distance to pivot. The tricky part here is working out the perpendicular distance as there is a temptation to try to use \(L\over 2\) - it's not impossible, but you would need to use trigonometry and eventually cancel side lengths.
HINT: On an exam paper, extend the line of action of the force (in this case vertically upwards and downwards) and then you will quickly spot the perpendicular distance.
A ladder leans against a wall.
What is the resultant torque about the top?
The ladder is in equilibrium so torques are balanced. There is no resultant torque.
A ladder leans against a wall.
Which equation is derived from balancing the torques about the top?
Notice that two torques are clockwise: \(W{b\over 2} + Fa\)
One torque is anticlockwise: \(N_bb\)
Which force causes an anticlockwise torque about the centre?
\(F\) and \(N_t \) cause clockwise torques.
\(W\) does not cause a torque about the centre as there is no perpendicular distance.
The image represents a force acting on identical rods, each with an identical mass attached.
Which arrangement will have the largest angular acceleration about the right end?
A and D have the smallest \(I\) (due to the reduced distance of the mass from the pivot).
A and B have the largest \(\Gamma\) (due to the increased distance of the force from the pivot).
\(α = {T\over I}\)
The image represents a force acting on identical rods, each with an identical mass attached.
The ratio \(\alpha_A \over \alpha_B\) is
\(α = {T\over I}\)
\(α_A = {FL\over m({L\over 2})^2} = 4{F\over mL}\)
\(α_B = {F{L\over 2}\over {mL}^2} = {1\over 2}{F\over mL}\)
HINT: You could also do this in your head. The torque around A is twice the size. The moment of inertia of B is one quarter the size (due to the halved length).
The image represents a force acting on identical rods, each with an identical mass attached.
The ratio \(\alpha_A \over \alpha_B\) is
Same torque so \(\alpha \propto {1\over I}\)
\( I = mr^2\) so \(I_B = 4I_A\)
In this diagram of a rod pivoted at C, the torques are not balanced.
Which of the following changes would bring about equilibrium?
Initially, \(F_a\) and \(F_b\) and are equal and acting in opposite senses, but the distances are not equal. We need to double \(\Gamma_b\) (or halve \(\Gamma _a\)) and only one option will achieve this.
The balanced beam has mass = 800 kg.
The force exerted by the left hand support is
Taking torques about the right hand end:
\(N \times 4 = 8000 \times 3\) (choosing \(4\) and \(3\) as distances is equivalent to using \(200\) and \(150\))
\(N = 6000 \text{ N}\)
NOTE: We could also answer this question by inspection. The left hand support must take more than half, but not all, of the weight.
The balanced beam has mass = 800 kg.
The force exerted by the right hand support is
Taking torques about the left hand end:
\(F \times 4 = 8000 \times 1\)
\(F = 2000\text{ N}\)
NOTE: Having calculated the left support force, we could alternatively have used linear equilibrium with resultant force \(=0\), i.e. \(8000=6000 + S_r\)
How much of Equilibrium and angular acceleration have you understood?
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