ROTATIONAL KINETIC ENERGY

ROTATIONAL KINETIC ENERGY


1. object of a certain mass moving with particular speed will have an associated kinetic energy  mass x speed².
2. An object spinning about an axis will also have associated with it a kinetic energy, composed of the kinetic energies of each individual part of the object. 
3These individual contributions may be summed up to give an expression for the total kinetic energy of the spinning object:





3. As with linear motion, where a force did work on an object and led to a change in the object's kinetic energy, for rotational motion the work done by a torque:





4. changing the rotational kinetic energy of an object:







5. Example :
     
    A) Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of 1.30 m and a mass of 72.0 kg. To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3600 m/s^2.

What I did was solve for the angular velocity through the radial acceleration:

= 3600m/s^2 = rw^2 
= w = 52.6 rad/s

Then I solved for the moment of inertia:

I = mr^2 = 72.0kg(1.30m)^2 = 122 kg*m^2

Finally I plugged it all into the rotational kinetic energy equation:

K = (1/2)(122m*m^2)(52.6rad/s)^2 

    = 1.68*10^5 J

B) A uniform rod of mass 0.20kg and lenght 1.20m is hinged freely about a horizontal axis

     through one end of the rod. The rod is raised so that it is horizontal and then released from

     the rest so that it rotates about the axis in a vertical plane.

     Calculate the

     (a) angular velocity

     (b) angular acceleration 

     when the rod makes an angle of 60º to the vertical.

[Moment of inertia of a uniform rod about a perpendicular axis through one end of is 1/3 Ml^2 where M = mass of rod ,l = length]

Example:
  a) Calculate the kinetic energy of a rigid body with a mass of 100 kg which is travelling at 20 m/sec.
                           Solution:
                                   Kinetic energy E_k = (1/2) mv²
                            Mass = 100 kg and speed = 20 m/sec
                                           = (1/2)(100)(20²)
                                           = 20000.
                          The kinetic energy is expressed in joules
                                 Kinetic energy E_k = 20000 joules.

SOLUTION

a) Using the priciple of conservation of energy,

    1/2Iw^2 = mgh

    1/2 (1/3 MI^2) w^2 = Mg(1/2 sin 30º )

    w^2   = 3g sin30º 

                       l

              =3 x 9.81 x 0.5

                      1.20

    w^2   =3.50rad s^-1

b) Using torque =Ia and torque = Fr,   (a = angular acceleration)

    

              mg x 1/2 sin 60º = Ml^2a 

                                                 3

              Angular acceleration , a = 3g sin 60º

                                                                 2l

                                                         = 3 x 98.1 sin 60º

                                                                     2 x 1.20

                                                         = 10.6 rad s^-2

    

ROTATIONAL WORK

ROTATIONAL WORK


1. Work is force times displacement, so for rotation work must be torque times angular displacement:


      

           


Example 


1. Change revolutions to angle in radians.

     a. 0.5 rev =   3.14 rad  
     b. 5 rev =   31.4 rad 
     c. 3.2 rev =   20.1 rad  

2. When a torque causes a rotation through an angle, the size of the angle can be expressed in degrees or 
     radians  .


3. Work done by a torque is equal to the torque applied times the angle turned through in radians

4.

A flywheel requires 75 ft·lb of work to stop its motion when a braking force of 3 lb is applied at the outer edge. The diameter of the flywheel is 22 in. (Assume that the force is applied tangent to the flywheel or, in other words, that the force is perpendicular to the radius.)A. Find the applied torque in ft·lb. B. Find how many revolutions the flywheel makes as it stops.   Known W = 75 ft·lb F = 3 lb d = 22 in 1 ft = 12 in  1 rev = 2p rad Unknown A) t = ? in ft·lb B) q = ? in rev Equations/Remarks A) L = r = d/2     t = F · L = (F · d)/2     t = (3 lb)(22 in)(1 ft/12 in)/2     t = 2.75 ft·lbB) W = t · q     q = W/t     q = (75 ft·lb)/(2.75 ft·lb)     q = (27.27 rad)(1 rev/2p rad)     q = 4.34 rev

5. A 196 N force is applied to the end of a torque wrench. The arm of the torque wrench is 46 cm long. It causes the wrench to move through an angle of p/4 radians. Find the work done in joules.

  Known

 F = 196 N

L = 46 cm

q = p/4 rad

1 m = 100 cm

1 J = 1 N·m

 Unknown

 W = ? in J

Equations/Remarks

 W = t · q
 t = F · L

t = F · L

W = t · q = F · L · q

W = (196 N)(46 cm)(p/4 rad)(1 m/100 cm)

W = 70.8 J

ROTATIONAL WORK AND KINETIC ENERGY

                

1. MOHD DIN BIN ELSHAL                                                                      54140111069

2.HAMIM SYAKIR BIN SUWARDI                                                            54140111066

3.MUHAMMAD FARHAN BIN MOHD ZAIRY                                        54140111046

4.MOHAMAD NIZAM BIN NOOR MOHAMAD                                       54140111026

ROTATIONAL WORK AND ENERGY

Rotational Work

Work-Energy Principle

The work-energy principle is a general principle which can be applied specifically to rotating objects. For pure rotation, the net work is equal to the change in rotational kinetic energy:

For a constant torque, the work can be expressed as

and for a net torque, Newton's 2nd law for rotation gives

Combining this last expression with the work-energy principle gives a useful relationship for describing rotational motion.

1.

In the picture given above F pulls a box having 4kg mass from point A to B. If the friction constant between surface and box is 0,3; find the work done by F, work done by friction force and work done by resultant force.

Work done by F;

W_F=F.X=20.5=100 joule

Work done by friction force;

W_friction=-F_f.X=-k.mg.X=-0,3.4.10.5=-60 joule

Work done by resultant force;

W_net=F_net.X=(F-F_f).X=(20-0,3.4.10)5

W_net=40 joule

Kinetic Energy

Kinetic energy is the energy of motion. An object that has motion - whether it is vertical or horizontal motion - has kinetic energy. There are many forms of kinetic energy - vibrational (the energy due to vibrational motion), rotational (the energy due to rotational motion), and translational (the energy due to motion from one location to another). To keep matters simple, we will focus upon translational kinetic energy. The amount of translational kinetic energy (from here on, the phrase kinetic energy will refer to translational kinetic energy) that an object has depends upon two variables: the mass (m) of the object and the speed (v) of the object. The 

following equation is used to represent the kinetic energy (KE) of an object

• A rotating body posses kinetic energy, because its constituent particles are moving.
• If the body is rotating with an angular speed w, the tangential speed VT  of a particles at a distance r from the axis is VT = rw
• If the particle's mass is m, it's kinetis energy is 1/2 m(VT)2
• Finally we define KE of rational motion as :  
     
                        

                                                                     

Definition of rational KE:

the rational KE of a girid object rotating with an angular speed w about a fixed axis and having a moment of inertia I.

• Requirement ; w must be expressed in rad/s
• SI Unit ; Joule ( J )

The expressions for rotational and linear kinetic energy can be developed in a parallel mannerfrom the work-energy principle. Consider the following parallel between a constant torque exerted on a flywheel with moment of inertia I and a constant force exerted on a mass m, both starting from rest.

Kinetic Energy of Rolling Object

If an object is rolling without slipping, then its kinetic energy can be expressed as the sum of the translational kinetic energy of its center of mass plus the rotational kinetic energy about the center of mass. The angular velocity is of course related to the linear velocity of the center of mass, so the energy can be expressed in terms of either of them as the problem dictates, such as in the rolling of an object down an incline. Note that the moment of inertia used must be the moment of inertia about the center of mass. If it is known about some other axis, then the parallel axis theorem may be used to obtain the needed moment of inertia.