Download Torque

ENGINEERING PHYSICS SECTION B1: RIGID BODIES, ROTATIONAL DYNAMICS
Key ideas:
Torque and Couple
Moment of inertia
Rotational and translational equilibrium
Angular acceleration
Equations of rotational motion for uniform angular acceleration
Newton’s second law applied to angular motion
Conservation of angular momentum
You have to be able to calculate torque for single forces and couples
Solve problems involving moment of inertia, torque and angular acceleration
Solve problems in which objects are in both rotational and translational equilibrium
Solve problems using rotational quantities analogous to linear quantities – the rotational
equivalent of SUVAT
Sketch and interpret graphs of rotational motion
Solve problems involving rolling without slipping (analysis will be limited to basic geometric
shapes)
The equation for the moment of inertia of a specific shape will be provided when necessary
Graphs will be limited to angular displacement–time, angular velocity–time and torque–time
LINEAR
ANGULAR
Γ = 𝐹𝑟𝑠𝑖𝑛𝜃
I = mr2
Γ = 𝐼𝛼
𝜔 = 2𝜋𝑓
𝜔𝑓 = 𝜔𝑖 + 𝛼𝑡
𝜔𝑓 2 − 𝜔𝑖 2 = 2𝛼𝜃
1
𝜃 = 𝜔𝑖 𝑡 + 𝛼𝑡 2
2
𝐿 = 𝐼𝜔
1
𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝐸𝐾 = 𝐼𝜔2
2
v = u + at
𝑣 2 − 𝑢2 = 2𝑎𝑠
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
𝑝 = 𝑚𝑣
𝑇𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝐸𝐾 =
1
𝑚𝑣 2
2
𝑠
Recall definitions for angular displacement 𝜃 = 𝑟,
angular velocity 𝜔 𝑖𝑛 𝑟𝑎𝑑 𝑠 −1 =
Δ𝜃
Δ𝑡
=
𝑣
𝑟
and angular acceleration  (by analogy) =
𝑤ℎ𝑒𝑟𝑒 𝑠 = 𝑣Δ𝑡
Δ𝜔
Δ𝑡
(𝑟𝑎𝑑 𝑠 −2 )
Torque Γ (Nm not J since F and r do not act along the same straight line)
The turning effect of a force. Imagine a spanner or wrench tightening a bolt.
JOHN MACARTHUR
1
ENGINEERING PHYSICS SECTION B1: RIGID BODIES, ROTATIONAL DYNAMICS
If a force F of 120N is applied perpendicular to the lever arm at a distance r of 0.2m, then
torque Γ = Fr (you learned this in MYP as the ‘moment’ or ‘turning effect’ of the force)
More precisely, Γ = Frsin For max torque, 900 so sin 900 

In the second diagram, the torque is reduced because the effort force is no longer at 90 0 to
the lever.
Find the angle between the effort force and the lever.
Solution: sin(angle required) =15/20 , so sin-1 0.75 = 48.60.
Thus Γ= 120 x 0.15 = 18Nm
The longer the lever, the greater the torque it can exert.
The idea of a couple.
The torque (or turning effect) due to a couple is the force x perpendicular distance between
the two forces.
JOHN MACARTHUR
2
ENGINEERING PHYSICS SECTION B1: RIGID BODIES, ROTATIONAL DYNAMICS
The idea of equilibrium. (exam: “what is meant by…”)
An object in equilibrium has no net ‘influences’ to cause it to move in a straight line or to
rotate. In other words, the sum of all forces acting is zero and also the sum of all torques
about a point is also zero. IOW, it stays in a state of uniform motion in a straight line or is at
rest (N1) and is not rotating about an axis.
At an instant (i): if these two conditions are satisfied the object or system is in equilibrium.
Examples: force only: try these for revision
JOHN MACARTHUR
3
ENGINEERING PHYSICS SECTION B1: RIGID BODIES, ROTATIONAL DYNAMICS
Balancing torques
Remember: it is the sum of all the ACW torques which equals the sum of the CW torques.
Consider the example top left, sometimes called the Decorator Problem. Here it is,
expanded a bit.
A decorator stands on a plank held up by two ropes. In this case, the mass of the plank has
been neglected. If it wasn’t, the weight force of the plank would act at its mid-point as the
upper diagram shows and would itself produce a torque about A or B.
Clearly 𝑇1 > 𝑇2 because the decorator’s weight W is closer to T1
The trick is to take moments about A, (or indeed B, it doesn’t matter), then solve as shown,
as long as we know the numerical values of W, d1 and d2.
JOHN MACARTHUR
4
ENGINEERING PHYSICS SECTION B1: RIGID BODIES, ROTATIONAL DYNAMICS
Example: decorator weight
=800N, stands on a plank 4m
long, ropes 1m from each end.
Plank weight = 40N
Find T1, T2 if d1 = 0.5m
JOHN MACARTHUR
5