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Rotational Motion question

2014 · Shift 0 · Q70
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Rotational Motion question

2014 · Shift 0 · Q70

JEE MainPhysicsRotational MotionMCQ+4 / −1
A bob of mass mmm attached to an inextensible string of length lll is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s\omega \,rad/sωrad/s about the vertical. About the point of suspension:
  1. A
    angular momentum is conserved
  2. B
    angular momentum changes in magnitude but not in direction.
  3. C
    angular momentum changes in direction but not in magnitude.
  4. D
    angular momentum changes both in direction and magnitude.
View written solutionFree

Correct answer: C

  1. Identify the motion

The bob executes a conical pendulum motion. The string makes a constant angle θ\thetaθ with the vertical, and the bob moves in a horizontal circle of radius

r=lsin⁡θ.r = l\sin\theta.r=lsinθ.

Its angular speed about the vertical axis is ω\omegaω.

We need the behavior of the angular momentum of the bob about the point of suspension.

  1. Write the position and velocity vectors

Take the point of suspension as origin and the vertical axis as zzz-axis.

At any instant,

r⃗=lsin⁡θ ρ^−lcos⁡θ z^.\vec r = l\sin\theta\,\hat \rho - l\cos\theta\,\hat z.r=lsinθρ^​−lcosθz^.

Since the bob moves in a horizontal circle with angular speed ω\omegaω, its velocity is tangential:

v⃗=ωlsin⁡θ ϕ^.\vec v = \omega l\sin\theta\,\hat \phi.v=ωlsinθϕ^​.

So angular momentum about the point of suspension is

L⃗=mr⃗×v⃗.\vec L = m\vec r\times \vec v.L=mr×v.
  1. Compute L⃗\vec LL

Using

r⃗=lsin⁡θ ρ^−lcos⁡θ z^,v⃗=ωlsin⁡θ ϕ^,\vec r = l\sin\theta\,\hat \rho - l\cos\theta\,\hat z, \qquad \vec v = \omega l\sin\theta\,\hat \phi,r=lsinθρ^​−lcosθz^,v=ωlsinθϕ^​,

we get

L⃗=m(lsin⁡θ ρ^−lcos⁡θ z^)×(ωlsin⁡θ ϕ^).\vec L = m\left(l\sin\theta\,\hat \rho - l\cos\theta\,\hat z\right) \times \left(\omega l\sin\theta\,\hat \phi\right).L=m(lsinθρ^​−lcosθz^)×(ωlsinθϕ^​).

Now use the cylindrical unit-vector cross products:

ρ^×ϕ^=z^,z^×ϕ^=−ρ^.\hat\rho\times\hat\phi = \hat z, \qquad \hat z\times\hat\phi = -\hat\rho.ρ^​×ϕ^​=z^,z^×ϕ^​=−ρ^​.

Thus,

L⃗=mωl2sin⁡θ(sin⁡θ z^+cos⁡θ ρ^).\vec L = m\omega l^2\sin\theta\left(\sin\theta\,\hat z + \cos\theta\,\hat \rho\right).L=mωl2sinθ(sinθz^+cosθρ^​).

So,

L⃗=mωl2(sin⁡2θ z^+sin⁡θcos⁡θ ρ^).\vec L = m\omega l^2\left(\sin^2\theta\,\hat z + \sin\theta\cos\theta\,\hat\rho\right).L=mωl2(sin2θz^+sinθcosθρ^​).
  1. Check the magnitude of L⃗\vec LL

The angle between r⃗\vec rr and v⃗\vec vv is 90∘90^\circ90∘, so directly,

∣L⃗∣=mrv=m(l)(ωlsin⁡θ)=mωl2sin⁡θ.|\vec L| = mrv = m(l)(\omega l\sin\theta)=m\omega l^2\sin\theta.∣L∣=mrv=m(l)(ωlsinθ)=mωl2sinθ.

Since m,l,ω,θm,l,\omega,\thetam,l,ω,θ are constants in steady conical motion,

∣L⃗∣=constant.|\vec L| = \text{constant}.∣L∣=constant.

So the magnitude does not change.

  1. Check the direction of L⃗\vec LL

From

L⃗=mωl2(sin⁡2θ z^+sin⁡θcos⁡θ ρ^),\vec L = m\omega l^2\left(\sin^2\theta\,\hat z + \sin\theta\cos\theta\,\hat\rho\right),L=mωl2(sin2θz^+sinθcosθρ^​),

we see that z^\hat zz^ is fixed, but ρ^\hat\rhoρ^​ rotates as the bob goes around the circle.

Therefore, the direction of L⃗\vec LL changes continuously with time.

So:

  • magnitude of L⃗\vec LL = constant,
  • direction of L⃗\vec LL = changing.
  1. Torque check (for consistency)

The forces on the bob are:

  • tension along the string, which passes through the suspension point, so its torque about the suspension point is zero,
  • weight mg⃗m\vec gmg​, whose torque about the suspension point is nonzero.

Hence,

dL⃗dt=τ⃗≠0,\frac{d\vec L}{dt}=\vec\tau \neq 0,dtdL​=τ=0,

so angular momentum is not conserved. Since the torque is horizontal and perpendicular to the vertical component structure, it changes the direction of L⃗\vec LL, not its magnitude.

  1. Evaluate options
  • A: angular momentum is conserved — False
  • B: angular momentum changes in magnitude but not in direction — False
  • C: angular momentum changes in direction but not in magnitude — True
  • D: angular momentum changes both in direction and magnitude — False

Final Answer: C\boxed{\text{C}}C​

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