Torque

Student: Mohammed Bdair

Student ID # 0564224

Midterm Project

PH-260

12/13/13

Torque and Seesaw

A see-saw is an example of a simple lever. At the center of the see-saw, balancing either

side is the fulcrum . The fulcrum serves as an axis of rotation for the see-saw.

The two arms of the see-saw are called the lever arms. A lever arm is defined as the

distance between the axis of rotation and the line of action . The line of action is a line that

is perpendicular to the point at which the force is exerted.

Project Problem:

Two children, of unequal weight, to balance on a seesaw so that they will be able to make it

tilt back and forth without the heavier child simply sinking to the ground. Given that the

heavier child of weight (22 kg), is sitting a distance ( 1m) to the left of the pivot, the second

child of weight (20 kg) sits on the right side of the pivot at a distance (1 m) , What is the net

torque on the seesaw? Assume that the Board weighs ( 12kg) and (2.5m) long

Free body diagram

In the left side where the first child sits, we have force due to gravity (F1), and F2 on the

right side, In the middle at pivot point we have FgBoard due to the weight of the board

pointing downwards and Fn , the normal force pointing upwards.

Problem Discussion:

The sum of the torques ( net) equals :

= F1d1 + F2d2

= (22 kg)(9.8 m/s)(1 m) + (20 kg)(9.8 m/s)( -1 m)

= 215.6 N*m + -196 N*m = 19.6 N*m

We note that one torque will provide a rotational force in the

counterclockwise direction (F1) while the other force will provide a

rotational force in the clockwise direction (F2).

Fn and FgBoard

In this case no equilibrium is achieved and this case is an unbalanced see-

saw.

Equilibrium : Equilibrium exits when the seesaw is balanced such that it will not tend to rotate around the fulcrum.

At Equilibrium:

There is no net torque. There is no net force.

Conditions for Equilibrium The sum of the forces must equal zero. (Translational Equilibrium)

o F = F1 + F2 + FgBoard Fn = 0 ,

o F = m1g + m2g + mBoard g Fn = 0

The sum of the torques must equal zero. ( Rotational Equilibrium)

o = F1d1 + F2d2 = 0

Let's solve for the translational Equilibrium,

F = F1 + F2 Fn = 0

= (22 kg)(9.8 m/s) + (20 kg)(9.8 m/s) + (12 kg)(9.8 m/s) - Fn = 0

So Normal Force Pointing Upward should have this value,

Fn = 529.2 N.

We note that this doesn't necessarily mean static: static means that not only do the forces vanish but the velocities vanish as well.

The sum of the torques must equal zero. (Rotational Equilibrium)

o = F1d1 + F2d2 = 0

To achieve such an equilibrium we must have two equal forces or increase the distance from pivot point to the smaller force ( the second child in our case) such that the two torques are equal . Let's calculate the distance needed for the second child (d2) that balances the seesaw.

= (22 kg)(9.8 m/s)(1 m) + (20 kg)(9.8 m/s)( d2) = 0

d2 = 1.1 m .

Center of Mass : is the average position of the mass of the two children, neglecting the mass of the board.

xcm = [ (22kg)(1m) + (20kg)(1.1m) ]/[ (22 kg) + (20kg) ]

xcm = 1.0476 m

The center of Mass of the board (symmetric) lies on the axis of symmetry, which is the geometric center of the board.

Angular Acceleration ......... Unbalanced Case

When the see-saw acceleration.

The angular acceleration is inversely proportional to the analogy of the mass (moment of inertia " I" ) in a rotating system

Moment of Inertia for the board is given by formula:

Moment of Inertia for the system =

I=1/12(12kg) (2.5 m)^2+(22kg) (1m)^2 + (20kg) (1.1m)^2 I=52.45 kgm^2

Angular Acceleration : = /I

= 19.6 N*m /52.45 kgm^2 = 0.374 rad/ s

Angular Momentum

First we need to calculate Angular velocity.

When a board is first released ( 0 = 0) it rotates about the pivot point. If it takes 1.2

seconds to attain a value of angular velocity, we can determine the average angular

velocity of the system.

( t - 0 )

t = 0.374 rad/ s * 1.2 s

t = 0 .449 rad/ s

Angular Momentum: L = I

L = 52.45 kgm^2 * 0 .449 rad/ s

L = 23.539 kg m2 / s

conservation of Angular Momentum

If the seesaw is at rest, it will stay at rest. If it is rotating, it will keep on rotating ( Assuming Dissipative Friction forces = 0). Each body has a property called rotational inertia.

What we have here is, in fact, another conservation law. If the net torque is zero, then angular momentum is constant or conserved. We can see this by considering for the situation in which the net torque is zero.

Design and Material Consideration: See-saw can built from different types of materials such as wood, hard plastics or even metals , the significant factor related to the physics of see-saw is the control of torque on both sides of the board; i.e. length of the board must be average in order to produce an acceptable level of torque, the greater the amount of torque, the greater the tendency of the see-saw to be put into rotation. The overall design of a see-saw, in particular the fact that it sits on the ground, means that its board can never undergo anything close to 360 rotation; nonetheless, the board does rotate within relatively narrow parameters.

References:

1) College Physics, Volume 10 By Raymond A. Serway, Chris Vuille, Jerry S. Faughn.

P. 232-245

2) hyperphysics.phy-astr.gsu.edu/hbase/amom.html 3) www.physics.uoguelph.ca/tutorials/torque/Q.torque.intro.angacc.html

4) http://ffden-2.phys.uaf.edu/212_spring2011.web.dir/Samantha_Porreca/Seesaw.html

5) http://www.engineeringtoolbox.com/moment-inertia-torque-d_913.html 6)https://notendur.hi.is/eme1/skoli/edl_h05/masteringphysics/9_10/torquesonseesaw.htm