Question
Physics Question on rotational motion
Two discs of moment of inertia I1=4kgm2 and I2=2kgm2 about their central axes and normal to their planes, rotating with angular speeds 10rad/s and 4rad/s respectively are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is __________ J.
The kinetic energy of a rotating body is given by the formula:
KE=21Iω2,
where I is the moment of inertia and ω is the angular velocity.
Kinetic Energy of Each Disc Before Contact: For disc 1:
KE1=21I1ω12=21(4kg m2)(10rad/s)2=21×4×100=200J.
For disc 2:
KE2=21I2ω22=21(2kg m2)(4rad/s)2=21×2×16=16J.
Total Kinetic Energy Before Contact:
KEtotal initial=KE1+KE2=200J+16J=216J.
Finding Final Angular Velocity After Contact: When the discs come into contact, they will rotate together, and we can use the principle of conservation of angular momentum. Initial angular momentum Linitial:
Linitial=I1ω1+I2ω2=(4kg m2)(10rad/s)+(2kg m2)(4rad/s)=40+8=48kg m2/s.
The total moment of inertia after they are in contact:
Itotal=I1+I2=4kg m2+2kg m2=6kg m2.
Final angular velocity ωf:
Lfinal=Itotalωf⟹48=6ωf⟹ωf=8rad/s.
Final Kinetic Energy After Contact:
KEfinal=21Itotalωf2=21(6kg m2)(8rad/s)2=21×6×64=192J.
Calculating Loss in Kinetic Energy:
Loss in KE=KEtotal initial−KEfinal=216J−192J=24J.