Pptg101213 Rotations Gallego

• The meter stick has mass. • Each point of this meter stick has mass. • The entire mass of the meter stick is

represented by a single point called the “center of mass”.

The center of mass is the average location of the entire mass of the object such that, when that single point is supported, it is like supporting the whole object.

Suppose that one end of the meter stick hangs unsupported, until how much of it should be exposed before it starts to topple? Why?

When that center of mass is not supported, the meter stick will have the tendency to rotate.

Rotations • Difference between rotation and revolution.

• Toppling and stability

• Torque

Rotation vs Revolution

The disc rotates, the man revolves but they are both turning! Rotation means turning about own axis like Earth’s rotation. Revolution means turning around another object like Earth revolving around the sun.


When the center of gravity shifts off of the support, rotation takes place.


In the picture above, the CG is not located on the fulcrum but to the right side of the fulcrum.

Toppling and Stability

For as long as the CG is within the base, there is no rotation or no toppling.

Toppling and Stability

Where is the CG of this setup?

The center of gravity falls within the alignment of the base. Most probably in this setup it represented by the red dot.

Toppling & Stability

The centers of gravity of three trucks parked on a hill are shown by the X’s. Which truck(s) will tip over?

This car will tip because its center of gravity is outside of its base.


Three balls contain weight located in different positions are standing on a floor. Which ball would least like to tip over? Why?

Rotation is lesser when the CG is near the ground. Ball A would least like to tip over.


Stability is determined by the distance that the CG is raised in tipping. An object with a wider base and low CG is more stable.


Why won’t this topple?

CG is located below the support. Rotation is impossible!


Explain why a long pole is beneficial to a tight rope bicycle rider if the pole droops?

When the pole droops the CG will shift downward thus the bicycle rider will become more stable.


In order to be more stable, the CG has to be in located in the lowest point or in some instances below the support.


Unbalanced torque () causes rotation.

The closer the CG to the ground, the lower is the lever arm hence, torque is small.

Torque

Consider a weight on the center of a meter stick which you are holding and you would move the weight closer and father from your grip, what would you feel?

Does the force change when you move the weight?

Torque

Force

Tends to change the motion of things.

Torque

Tends to change the rotation of things.

Torque

Torque () = lever arm (r) x force (F) When the lever arm, r, is increased by moving the weight to the tip of the meter stick, the torque,, will increase.

: we denote lever arm as r to avoid confusion in the later part.

Torque

Torque () = lever arm (r) x force (F) When the lever arm, r, is increased by moving the weight to the tip of the meter stick, the torque,, will increase.

Torque

Torque () = lever arm (r) x force (F) When the lever arm,r, is increased by moving the weight to the tip of the meter stick, the torque,, will increase.

Torque

Torque () = lever arm (r) x force (F) When the force is applied an angle, only the component that is perpendicular to the lever arm counts.

=r Fsin

Torque (Problem Solving)

Calculate the torque produced by a 50-N perpendicular force at the end of a 0.2m-long wrench.

Calculate the torque produced by the same 50-N force when a pipe extends the length of the wrench to 0.5m.

τ = r F

= 0.2m x 50N

= 10N m

τ = r F

= 0.5m x 50N

= 25N m


Calculate the individual torques produced by the weights of the girl and boy on the seesaw in the figure. What is the net torque?

10m

300N 450N

τ = r F

= 5m x 300N

= 1,500N m

girl

τ = r F

= 5m x 450N

= 2,250N m

boy

Since torque is a vector quantity, direction of rotation matters. By convention, a negative torque tends to create a clockwise rotation.

τ = τ + τ

= 1,500N m + (-2,250N m)

= -750N m

net girl boy


Which of them needs an additional weight to make the seesaw balance?

10m

300N 450N

To make it balanced, an additional torque of 750N-m (from previous slide) in needed on the side of the girl.


Suppose that the available weight was only 300N, what distance should it sit from the fulcrum to make it balance?

10m

300N 450N

300

τ = r F

τr=

F750N m

= 300N

2.5m

300

It has to be placed at a distance 2.5m from the fulcrum.

Torque

A has torque and B has torque. When the torque of each is not equal, then they are no balanced. The seesaw will rotate in favor of the greater torque.

A B

Equilibrium

A B

= 0 = A+ B

.

A = B Lever armA x Force A = Lever armA x Force A

Equilibrium

A B

= 0 = A+ B

A = B

3m x 250N = 1.5mx 500 750N-m = 750N-m

Equilibrium

The mass of the rock is 1kg. What is the mass of the stick and where is the CG of the stick located?

The torque on each side should cancel each other or of the same magnitude but opposite in direction. Thus, the torque that a 1-kg rock can make is balanced by the torque of the stick. Since the CG of the stick and the rock is equidistant from the fulcrum, the mass of the stick is 1 kg and

the CG is located on its geometric center.

Equilibrium

= 0 = w+ cw

.

.

w = cw

What is this for?

This provides a torque that is

opposite to that of the load so no

rotation will occur. Hence it is called counter weight.

Equilibrium (Problem Solving)

50cm 75cm 100cm 25cm 0cm

1 kg

What is the mass of the stone?

τ = τ + τ

τ = τ

r F r F

r m g r m g

r mm

r

ball box

ball box

ball ball box box

ball ball box box

box boxball

ball

r mm

r

0.5m×1kg

0.25m

2 kg

box boxball

ball

Note: The r is the lever arm distance from the fulcrum.


If a 200-g mass is placed at the 20cm mark (30cm from the fulcrum) at what mark should a 500-g mass be placed so that

the system balances?

500

500

τ = τ

r F r F

r m g r m g

r m r

m

30cm×200gr =

500g

r = 12cm

200 500

200 200 500 500

200 200 500 500

200 200500

500

Since the case is in equilibrium, then, torques on both sides cancel each other.

Notice that g is cancelled and the unit of m need not to be in kilograms for as long as both m’s have the same units. They can be cancelled anyway.

Take note also that r is measured from the fulcrum, thus the 500g will be placed in 62cm mark of the meter stick.


If a 100-g mass was placed at the 25-cm mark, and a 20-g mass at the 10-cm mark, where should a 500-g mass be

placed to balance the system?

τ + τ = τ

r F + r F r F

r m + r m r

m

100 20 500

100 100 20 20 500 500

100 100 20 20500

500

500

500

500

500

25cm×100g + 40cm×20gr =

500g

2,500cm-g + 8,000cm-gr =

500g

10,500cm-gr =

500g

r = 21cm

100-x x


A horizontal meter stick has a 3-kg blob of clay stuck on one end at the 0 mark, and 1-kg blob stuck on the other end, at

the 100-cm mark. Neglecting the mass of the stick it self, where is the center of gravity of stick and clay system?

3kg 1kg

3kg 3kg 1kg 1kg

τ = τ

r ×F = r ×F

x×3kg= 100-x ×1kg

3x= 100-x

3x + x = 100

4x = 100

100x=

4x=25cm


A rigid plank of length L has 3 equal masses placed on it, one at each end and the 3rd is 2/3 of the length along the plank. What distance from the end of the plank should the pivot be

placed in order to attain a balance?

Lever

Is it possible to obtain a bigger torque with lesser force? How?

How does lever uses the concept of torque? How then, is torque related to work?

Pptg101213 Rotations Gallego

Documents

Transcript of Pptg101213 Rotations Gallego