Question

A cylinder of mass 11.0 kg and radius 0.260m rolls without slipping on a horizontal surface. At a particular instant, its center of mass has a speed of 7.30 m/s. Its rotational kinetic energy about its center of mass is then 147J.

a) What is in kg m² its moment of inertia about its center of mass?

b) What is in J its linear kinetic energy at that instant?

c) What is in J its total kinetic energy at that instant?

d) What is in kg m/s its linear momentum at that instant?

e) What is in kg m²/s its angular momentum about its center of mass at that instant?

Answer 1

Rotational kinetic energy is given as,

$E_{R}=0.5I\omega^{2}$

we know the relation between linear and rotational velocity as,

$v=r\omega\Rightarrow \omega =7.3/0.26=28.07rad/s$

(A) Now, moment of inertia I is,

$I=2E_{R}/\omega^{2}\Rightarrow I =0.373kg m^{2}$

(B) Linear kinetic energy is given as,

K.E.= 0.5mv^2= 293.095J

(C) Total kinetic energy is sum of roational and linear,

Total K.E= 147+239.095=440.095J

(D) Linear momentum is given as,

M= mv= 11*7.3= 80.3kgm/s

(E) Angular momentum is given as,

L=mrv= 11*0.26*7.3=20.88kgm^2/s

Answer 2

1/2*inertia*w^2=147

==>inertia=147*r^2*2/v^2

==>inertia=0.37294 kg m²

Kinetic energy=1/2*mass*v²

                       =11*7.3*7.3/2

                         =293.095 j

Total kinectic enetgy=293.095+147

                                  =440.095 j

Linear momentum=m*v=11*7.3=80.3kgm/s