Unit II: Physics of Extended Bodies and Thermodynamics
Rotational Motion
Recalling:
Vector- a physical quantity with magnitude and direction (i.e., 4m, East; 3km/h, North; 5m/s, West)
Mechanics- the study of motion.
Kinematics- the mathematical description of motion.
Dynamics- the study of causes of motion.
Rotational Kinematics
Rigid body- an extended object (not a point particle) with a definite shape.
- a body that does not deform or change shape during the course of motion (ie., ball bearing)
Angle θ can be measured in degrees or radians, but for rotational motion, it is simpler to measure angles
in radians than in degrees.
Radian- (abv.- rad) the angle subtended at the center of a circle by an arc with length s equal to the radius r
of the circle (refer to figure 9.3 below).
Generally, angle θ in radians is equal to s (arc) divided by r (radius) of the circle. Mathematically,
One full circle has an angle of 360º, corresponding to a circumference of 2π. Thus, 1 complete circle,
2
𝜋𝑟
𝑟
= 2π rad
For one revolution in a circle, you can write 1 rev = 360º = 2π rad.
Example 1:
Translational motion- movement of object from one point to another through space.
Angular displacement can be defined through: ∆θ = 𝜃2 - 𝜃1
Angular velocity can be defined through: 𝜔𝑎𝑣𝑒 =
∆𝜃
∆𝑡
=
𝜃 − 𝜃0
𝑡 − 𝑡0
Angular acceleration (α) is defined as change in angular velocity per unit time. Mathematically,
𝛼𝑎𝑣𝑒 =
Linear velocity can be defined through: v =
Example 2:
s
∆t
=r
∆𝜔
𝜔 − 𝜔0
=
∆𝑡
𝑡 − 𝑡0
∆θ
∆t
= rω
Radial acceleration- linear acceleration directed toward the center of the circle.
Angular acceleration- an angular quantity that describes how fast or slow the speed of the rotation changes
and whose direction is given by the right-hand rule.
Example 3:
Example:
Rotational Dynamics- the causes of rotational motion.
Torque- is the effort or cause for an object to rotate (i.e., you twist the cap of bottle to open it; when driving,
you turn the steering wheel to turn left or right).
- with the measure of how much force acting on an object causing it to rotate.
Example:
The object rotates about an axis, which we will call the pivot point, and will label 'O'. For the force, we call
it ‘F’ and the distance from the pivot point to the point where the force acts is called the lever arm, denoted
as ‘r’ (refer to the figure below).
Torque is mathematically defined as, τ = r x F
In other words, torque is the cross product between the radius ‘r’ vector and the force vector, 'θ' being the
angle between r and F.
Consider a screw shown in the diagram. If forces are applied at different locations then how the rotating
effects are created? The axis of rotation passes through the center of the screw.
First, consider only F1 force, will it create any rotational effect? The screw will not rotate, thus the force
F1 will not create any torque.
Now, if only force F2 is applied, then again the screw will not rotate.
Which force will create a torque? Forces F3 and F4 will rotate the screw, thus they will create some torque
or turning effect.
Which force can easily rotate the screw? Force F3 is far of the axis of rotation and will easily rotate the
screw as compared to F4..
You can see, F1 and F3 have the same distance from the axis of rotation, but different angles. Therefore,
torque depends on the orientation of force. Torque also depends on the magnitude of the force. The greater
the magnitude of force, the greater is the turning effect. Suppose you want to open a jammed screw, then
you have to apply greater force to produce greater torque.
Example 1:
Example 2:
Example 3:
Torque
Direction
Type of Rotation
a) 30 J, In the Plane of the Paper, Clockwise
Solutiom:
T=rxF
T = (3 m) (10 N)
T = 30 J
b) 18 J, Out the Plane of the Paper, Counterclockwise
Solutiom:
T=rxF
T = (3 m) (10 N) (sin(37º))
T = 18 J
c) 23 J, Out of the Plane of the Paper, Counterclockwise
Solutiom:
𝜏1 = r x F
𝜏1 = (1 m) (5 N)
𝜏1 = 5 J
𝜏2 = r x F
𝜏2 = (3 m) (10 N) (sin(37º))
𝜏2 = 18 J
𝜏𝑡 = 𝜏1 + 𝜏2
𝜏𝑡 = (5 J) + (18 J)
𝜏𝑡 = 23 J
d) : 21 J, Inside the Plane of the Paper, Clockwise
Solutiom:
𝜏1 = r x F
𝜏1 = (1 m) (5 N)
𝜏1 = 5 J
𝜏𝑡 = 𝜏1 + 𝜏2
𝜏𝑡 = (5 J) + (16 J)
𝜏𝑡 = 21 J
𝜏2 = r x F
𝜏2 = (2 m) (10 N) (cos(37º))
𝜏2 = 16 J