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Question

Physics Question on rotational motion

Two discs of moment of inertia I1=4kgm2I_1 = 4 \, kg \, m^2 and I2=2kgm2I_2 = 2 \, kg \, m^2 about their central axes and normal to their planes, rotating with angular speeds 10rad/s10 \, rad/s and 4rad/s4 \, rad/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.

Answer

The kinetic energy of a rotating body is given by the formula:

KE=12Iω2KE = \frac{1}{2} I \omega^2,

where II is the moment of inertia and ω\omega is the angular velocity.

Kinetic Energy of Each Disc Before Contact: For disc 1:

KE1=12I1ω12=12(4kg m2)(10rad/s)2=12×4×100=200JKE_1 = \frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} (4 \, \text{kg m}^2)(10 \, \text{rad/s})^2 = \frac{1}{2} \times 4 \times 100 = 200 \, \text{J}.

For disc 2:

KE2=12I2ω22=12(2kg m2)(4rad/s)2=12×2×16=16JKE_2 = \frac{1}{2} I_2 \omega_2^2 = \frac{1}{2} (2 \, \text{kg m}^2)(4 \, \text{rad/s})^2 = \frac{1}{2} \times 2 \times 16 = 16 \, \text{J}.

Total Kinetic Energy Before Contact:

KEtotal initial=KE1+KE2=200J+16J=216JKE_{\text{total initial}} = KE_1 + KE_2 = 200 \, \text{J} + 16 \, \text{J} = 216 \, \text{J}.

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 LinitialL_{\text{initial}}:

Linitial=I1ω1+I2ω2=(4kg m2)(10rad/s)+(2kg m2)(4rad/s)=40+8=48kg m2/sL_{\text{initial}} = I_1 \omega_1 + I_2 \omega_2 = (4 \, \text{kg m}^2)(10 \, \text{rad/s}) + (2 \, \text{kg m}^2)(4 \, \text{rad/s}) = 40 + 8 = 48 \, \text{kg m}^2/\text{s}.

The total moment of inertia after they are in contact:

Itotal=I1+I2=4kg m2+2kg m2=6kg m2I_{\text{total}} = I_1 + I_2 = 4 \, \text{kg m}^2 + 2 \, \text{kg m}^2 = 6 \, \text{kg m}^2.

Final angular velocity ωf\omega_f:

Lfinal=Itotalωf    48=6ωf    ωf=8rad/sL_{\text{final}} = I_{\text{total}} \omega_f \implies 48 = 6 \omega_f \implies \omega_f = 8 \, \text{rad/s}.

Final Kinetic Energy After Contact:

KEfinal=12Itotalωf2=12(6kg m2)(8rad/s)2=12×6×64=192JKE_{\text{final}} = \frac{1}{2} I_{\text{total}} \omega_f^2 = \frac{1}{2} (6 \, \text{kg m}^2)(8 \, \text{rad/s})^2 = \frac{1}{2} \times 6 \times 64 = 192 \, \text{J}.

Calculating Loss in Kinetic Energy:

Loss in KE=KEtotal initialKEfinal=216J192J=24J\text{Loss in KE} = KE_{\text{total initial}} - KE_{\text{final}} = 216 \, \text{J} - 192 \, \text{J} = 24 \, \text{J}.