Answer:
a) 6.0 J
b) 2.0 J
Explanation:
The kinetic energy in this case is given by:
[tex]K=\frac{1}{2}*I*\omega^2[/tex]
where I is the moment of inertia, for both cases the moment of inertia is different.
For case A the moment of inertia of a particle is given by:
[tex]I=M*r^2[/tex]
and for B
[tex]I=\frac{1}{3}M*L^2[/tex]
Solving this:
[tex]K_a=\frac{1}{2}*(0.75kg*(0.87m)^2)*(4.6rad/s)^2\\\\K_a=6.0J\\\\K_a=\frac{1}{2}*(\frac{1}{3}*0.75kg*(0.87m)^2)*(4.6rad/s)^2\\\\K_a=2.0J[/tex]
(a) The kinetic energy of rod A is 6.0 J
(b) The kinetic energy of rod B is 2.0 J
The kinetic energy in rotational motion is given by:
[tex]KE=\frac{1}{2}I\omega^2[/tex]
where I is the moment of inertia, and ω is the angular velocity.
The moment of inertia is defined as:
[tex]I=MR^2[/tex]
where M is the mass and R is the distance from the axis of rotation.
In case of rod A :
the rod is massless so only the moment of inertial of the particle is considered, then
[tex]K_A=\frac{1}{2}MR^2\omega^2\\\\K_A=0.5\times0.75\times(0.87)^2\times(4.6)^2J\\\\K_A=6J[/tex]
In case of rod B:
The rod has a mass of 0.75 kg uniformly distributed and the moment of inertia of a uniform rod is given by;
[tex]I=\frac{1}{3}ML^2[/tex]
where L is the length of the rod
so,
[tex]K_B=\frac{1}{2}\times\frac{1}{3}\times0.75\times(0.87)^2\times(4.6)^2J\\\\K_B=2J[/tex]
Learn more about rotational motion:
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