POST 2

question 1

A lorry haning a mass of 1.5t is travel-ling along a level road at 72 km/h. when the brakes are applied, the speed decreases to 18 km/h. determine how much the kinetic enegy of the lorry is reduced.

Initial velocity of lorry, v1                  = 72 km/h

                                       km                    m            1h

                            =  72   −−−   x  1000     −−−    x    −−−

                                        H                     km         3600s

                                    72

                               =  −−−   = 20 m/s

                                   3.6

Final velocity of lorry, v2                =  18

                                                        −−−  =   5 m/s and

                                                         3.6

Mass of lorry , m                             = 1.5t = 1500 kg

Initial kinetic energy of the lorry        1                    1

                                                  = −−− mv ²   =   −−−  (1500)(20) ²

                                                      2          ¹        2                   

                                              

             = 300 kj

Final kinetic energy of the lorry         1                  1

                                                 =  −−  mv²  =   −− ( 1500)(5) ²

                                                      2     ²           2

                                                    

 = 18.75 kj

Hence, the change in kinetic energy    = 300 – 18.75

                            

                                                        = 281.25 kj

POST 2

A rod of mass  M=3KG  and length  L=1.2m  pivots about an axis, perpendicular to its length, which passes through one of its ends. What is the moment of inertia of the rod? Given that the rod's instantaneous angular velocity is  60deg/s  what is its rotational kinetic energy?

Answer: The moment of inertia of a rod of mass M and length L about an axis, perpendicular to its length, which passes through its midpoint is I = (1/12)ML². This is a standard result. Using the parallel axis theorem, the moment of inertia about a parallel axis passing through one of the ends of the rod is

I¹ I + M ( L / 2) ²=1/3 ML²

I¹ = 3 x 1.2² =1.44 kg m²

         −−−−−

            3

                                 

The instantaneous angular velocity of the rod is 

w = 60 x  π  = 1.047 rad/s

           −−−

            180

Hence, the rod's rotational kinetic energy is written 

K = 1 I¹ w² = 0.5 x 1.44 x 1.047² = 0.789 J

        −

        2