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JEE Physics 2024 Question with Solution

Two discs with moments of inertia I1=4 kg⋅m2I_1 = 4 \, \text{kg}\cdot\text{m}^2 and I2=2 kg⋅m2I_2 = 2 \, \text{kg}\cdot\text{m}^2 about their central axes, rotating with angular speeds 10 rad/s10 \, \text{rad/s} and 4 rad/s4 \, \text{rad/s}, respectively, are brought into contact face-to-face. The loss in kinetic energy of the system is:

  • A

    20 J20 \, \text{J}

  • B

    24 J24 \, \text{J}

  • C

    30 J30 \, \text{J}

  • D

    18 J18 \, \text{J}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: I1=4 kg⋅m2I_1 = 4 \, \text{kg}\cdot\text{m}^2, I2=2 kg⋅m2I_2 = 2 \, \text{kg}\cdot\text{m}^2, ω1=10 rad/s\omega_1 = 10 \, \text{rad/s}, and ω2=4 rad/s\omega_2 = 4 \, \text{rad/s}.

Find: the loss in kinetic energy when both discs rotate together after contact.

The kinetic energy of a rotating body is

KE=12Iω2KE = \frac{1}{2} I \omega^2

So, before contact:

KE1=12I1ω12=12(4)(10)2=200 JKE_1 = \frac{1}{2} I_1 \omega_1^2 = \frac{1}{2}(4)(10)^2 = 200 \, \text{J} KE2=12I2ω22=12(2)(4)2=16 JKE_2 = \frac{1}{2} I_2 \omega_2^2 = \frac{1}{2}(2)(4)^2 = 16 \, \text{J}

Hence, total initial kinetic energy is

KEinitial=200+16=216 JKE_{\text{initial}} = 200 + 16 = 216 \, \text{J}

Using conservation of angular momentum:

I1ω1+I2ω2=(I1+I2)ωfI_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2)\omega_f (4)(10)+(2)(4)=(4+2)ωf(4)(10) + (2)(4) = (4+2)\omega_f 48=6ωf48 = 6\omega_f ωf=8 rad/s\omega_f = 8 \, \text{rad/s}

Now the final kinetic energy is

KEfinal=12(I1+I2)ωf2=12(6)(8)2=192 JKE_{\text{final}} = \frac{1}{2}(I_1 + I_2)\omega_f^2 = \frac{1}{2}(6)(8)^2 = 192 \, \text{J}

Therefore, the loss in kinetic energy is

ΔK=KEinitial−KEfinal=216−192=24 J\Delta K = KE_{\text{initial}} - KE_{\text{final}} = 216 - 192 = 24 \, \text{J}

Therefore, the correct option is B.

This is an inelastic rotational interaction: angular momentum is conserved, but kinetic energy is not conserved.

Step-by-Step Working

Given: two discs are brought into contact face-to-face and finally rotate together.

Find: loss in kinetic energy of the system.

Step 1: Calculate initial angular momentum

Li=I1ω1+I2ω2L_i = I_1\omega_1 + I_2\omega_2 Li=(4)(10)+(2)(4)=40+8=48 kg⋅m2/sL_i = (4)(10) + (2)(4) = 40 + 8 = 48 \, \text{kg}\cdot\text{m}^2/\text{s}

Step 2: Calculate final common angular speed Total moment of inertia after contact is

If=I1+I2=4+2=6 kg⋅m2I_f = I_1 + I_2 = 4 + 2 = 6 \, \text{kg}\cdot\text{m}^2

Using Li=LfL_i = L_f,

48=6ωf48 = 6\omega_f ωf=486=8 rad/s\omega_f = \frac{48}{6} = 8 \, \text{rad/s}

Step 3: Calculate total initial kinetic energy

Ki=12I1ω12+12I2ω22K_i = \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2 Ki=12(4)(10)2+12(2)(4)2K_i = \frac{1}{2}(4)(10)^2 + \frac{1}{2}(2)(4)^2 Ki=2(100)+1(16)=200+16=216 JK_i = 2(100) + 1(16) = 200 + 16 = 216 \, \text{J}

Step 4: Calculate final kinetic energy

Kf=12(I1+I2)ωf2K_f = \frac{1}{2}(I_1 + I_2)\omega_f^2 Kf=12(6)(8)2=3(64)=192 JK_f = \frac{1}{2}(6)(8)^2 = 3(64) = 192 \, \text{J}

Step 5: Compute the loss

ΔK=Ki−Kf=216 J−192 J\Delta K = K_i - K_f = 216 \, \text{J} - 192 \, \text{J} ΔK=24 J\Delta K = 24 \, \text{J}

So, the loss in kinetic energy of the system is 24 J24 \, \text{J}, hence the correct option is B.

Common mistakes

  • Using conservation of kinetic energy is incorrect here because friction between the discs makes the rotational contact effectively inelastic. Use conservation of angular momentum, then compute the kinetic energy loss separately.

  • Taking the final moment of inertia as only one disc is wrong because after contact both discs rotate together. Use I1+I2I_1 + I_2 for the final combined system.

  • Subtracting angular speeds directly to estimate energy loss is incorrect because kinetic energy depends on ω2\omega^2, not linearly on angular speed. First find the common final angular speed from angular momentum conservation.

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