Physics 101: Lecture 17 Rotational Dynamics

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# Physics 101: Lecture 17 Rotational Dynamics - PowerPoint PPT Presentation

Physics 101: Lecture 17 Rotational Dynamics. Review of the last lecture on Rotational Kinematics and Rolling Motion and the Vector Nature of Angular Variables Today’s lecture will cover Textbook Sections 9.1 - 9.3. But first for something completely different ….

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Physics 101: Lecture 17Rotational Dynamics
• Review of the last lecture on Rotational Kinematics and

Rolling Motion and the Vector Nature of Angular Variables

• Today’s lecture will cover Textbook Sections 9.1 - 9.3
But first for something completely different …

The Quantum and the Cosmos

By Rocky Kolb

Today, 4:30 pm,

114 Hochstetter

Dark matter ?

How did our universe evolve out of a formless,

shapeless quantum soup of “elementary” particles ?

Dark Energy ?

See text: chapter 8

And for a point at a distance R from the rotation axis:

• x = Rv = R a = R

Rotation Summary (with comparison to 1-D linear kinematics)

Angular Linear

See Table 8.1

Rolling Motion
• Rolling motion involves both linear and rotational motion.

If there is no slipping at the point of contact with

the ground, the motion of an automobile tire is an

example for rolling motion.

While the car is moving with a linear speed v

covering a distance d, a point on the outer edge of

the tire moves the same distance along the circular

path s = q r = d.

Thus, the linear speed v=d/t of the car is the same

as the tangential speed of a point on the outer edge

on the tire vT=s/t=q/t r = w r:

v = vT = w r (w in rad/s)

Vector Nature of Angular Variables
• Like linear velocity and acceleration, also

angular velocity and acceleration are vector

quantities.

So far we only talked about the magnitude

of these vectors. But as vectors they also

have a direction. Both angular velocity and

acceleration point along the rotation axis.

As in linear motion, if the velocity is

decreasing, the acceleration points opposite

to the angular velocity and if the velocity is

increasing, velocity and acceleration point

in the same direction.

Rotational Dynamics
• Remember Newtons first and second laws:

An object does not change its state of motion unless

a net force is applied. If a net force is applied on an

object with mass m it experiences an acceleration:

a = Fnet/m or Fnet=m a

What is the equivalent of force and mass (inertia) in

rotational motion of a rigid object ?

See text: chapter 9

New Concept: Torque

• Torque is the rotational analog of force:
• Torque = (magnitude of force) x (lever arm)
• t = F l SI unit: [N m]

The lever arm is the distance between the axis of

rotation and the “line of action”, i.e. the line drawn

along the direction of the force.

The torque is positive, if resulting into a counter

clockwise rotation and negative for a

clockwise rotation.

CORRECT

Concept Question

The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case.

In which of the cases is the torque on the nut the biggest?

1. Case 1 2. Case 2 3. Case 3

Center of gravity

pivot

d

W=-mg

Rigid Objects in Equilibrium

A system is in equilibrium if and only if:

• a= 0  SFext = 0
•  = 0  S ext = 0 (about any axis)
• Torque about pivot due to
• gravity:
• g = m g d
• (gravity acts at center of gravity)

This object is NOT in equilibrium.

Center of Gravity
• The center of gravity of a rigid body is the point at

which the gravitational force is considered to act when

the torque due to gravity (weight) is calculated.

If an object is of symmetric shape and the weight is

distributed uniformly, the center of gravity is located at

the geometrical center of the object.

If the object is made of a group of objects with known

weights, then its center of gravity is determined

by the sum of the torques around an arbitrary axis:

xcg =(W1 x1 + W2 x2 +W3 x3 + …)/(W1+W2+W3+..)

xcg is the point at which the weight W1+W2+W3+… acts.

pivot

Center of gravity

pivot

d

Center of gravity

W=-mg

Not in equilibrium

Equilibrium

A method to find center of gravity of an irregular object

Torque
• Painter on plank (uniform) which sits on two supports.
• Free body diagram for plank:

FA

FB

-Mg

-mg

To compute the net torque, choose a rotation axis, e.g. let’s consider the

pivot is located at support A.

Torque
• If its just balancing on “B”, then FA = 0
• the only forces on the beam are:

x

FB

Mg

mg

Using FTOT = 0: FB = Mg + mg This does not tell us x

Torque
• Find net torque around pivot B: (or any other place)

FB

d1

d2

Mg

mg

t(FB ) = 0 since lever arm is 0

t(Mg) = Mgd1

Total torque = 0 = Mgd1-mgd2

t(mg) = -mgd2

So d2 = Md1 /mand you can use d1 to find x