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NVAMediumJEE 2024Torque & Angular Momentum

JEE Physics 2024 Question with Solution

A body of mass 5 kg5 \, \text{kg} moving with a uniform speed 32 m/s3\sqrt{2} \, \text{m/s} in the X−YX-Y plane along the line y=x+4y = x + 4. The angular momentum of the particle about the origin will be kg⋅m2/s\text{kg}\cdot\text{m}^2/\text{s}:

Answer

Correct answer:60

Step-by-step solution

Standard Method

Given: Particle of mass m=5 kgm = 5 \, \text{kg} moves along the line y=x+4y = x + 4 with speed 32 m/s3\sqrt{2} \, \text{m/s}.

Find: The angular momentum of the particle about the origin.

Any point on the line is (x,x+4)(x, x+4), so the position vector is

r⃗=x i^+(x+4) j^\vec r = x\,\hat i + (x+4)\,\hat j

The particle moves along the line y=x+4y = x + 4, whose direction vector is i^+j^\hat i + \hat j. Its unit vector is

12(i^+j^)\frac{1}{\sqrt{2}}(\hat i + \hat j)

Hence the velocity is

v⃗=32⋅12(i^+j^)=3i^+3j^\vec v = 3\sqrt{2} \cdot \frac{1}{\sqrt{2}}(\hat i + \hat j) = 3\hat i + 3\hat j

Using angular momentum,

L⃗=m(r⃗×v⃗)\vec L = m(\vec r \times \vec v)

Now,

r⃗×v⃗=∣i^j^k^xx+40330∣=(3x−3(x+4))k^=−12k^\vec r \times \vec v = \begin{vmatrix} \hat i & \hat j & \hat k \\ x & x+4 & 0 \\ 3 & 3 & 0 \end{vmatrix} = \big(3x - 3(x+4)\big)\hat k = -12\hat k

Therefore,

L⃗=5(−12)k^=−60k^  kg m2s−1\vec L = 5(-12)\hat k = -60\hat k \; \text{kg m}^2\text{s}^{-1}

So the magnitude is

∣L⃗∣=60  kg m2s−1|\vec L| = 60 \; \text{kg m}^2\text{s}^{-1}

Therefore, the angular momentum of the particle about the origin is 6060.

Perpendicular Distance Method

Given: Mass m=5 kgm = 5 \, \text{kg}, speed v=32 m/sv = 3\sqrt{2} \, \text{m/s}, path y=x+4y = x + 4.

Find: Magnitude of angular momentum about the origin.

For motion along a straight line, the magnitude of angular momentum about the origin is

L=mvdL = mvd

where dd is the perpendicular distance of the origin from the line of motion.

The line y=x+4y = x + 4 can be written as

x−y+4=0x - y + 4 = 0

So the perpendicular distance of the origin from this line is

d=∣4∣12+(−1)2=42=22d = \frac{|4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}

Now,

L=mvd=5×32×22=5×3×4=60L = mvd = 5 \times 3\sqrt{2} \times 2\sqrt{2} = 5 \times 3 \times 4 = 60

Therefore, the angular momentum is 6060.

Common mistakes

  • Using L⃗=Iω\vec L = I\omega here is incorrect because the particle is not given to be rotating about a fixed axis. Use L⃗=r⃗×mv⃗\vec L = \vec r \times m\vec v or L=mvdL = mvd for a moving particle.

  • Taking the distance from the origin as the distance of a point on the line is wrong. Angular momentum magnitude depends on the perpendicular distance from the origin to the line of motion, not the distance to an arbitrary point.

  • Ignoring direction and directly writing −60-60 as the final numerical answer is a mistake. The vector angular momentum is −60k^-60\hat k, but the required numerical value is its magnitude, which is 6060.

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