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Center of Mass question

2020 · 3 Sep · Shift 1 · Q51
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Center of Mass question

2020 · 3 Sep · Shift 1 · Q51

JEE MainPhysicsCenter of MassMCQ+4 / −1
A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle θ\thetaθ before momentarily coming to rest. If the rod has mass M = 2 kg, and length lll= 1 m, the value of θ\thetaθ is approximately : (take g = 10 m/s2) JEE Main 2020 (Online) 3rd September Morning Slot Physics - Center of Mass and Collision Question 72 English
  1. A
    63o
  2. B
    69o
  3. C
    55o
  4. D
    49
View written solutionFree

Correct answer: A

  1. Identify the physics involved

The collision is perfectly inelastic because the block sticks to the rod.
During the short collision, angular momentum about the pivot OOO is conserved.

After collision, the system rotates upward and comes momentarily to rest at angle θ\thetaθ. Then:

  • rotational kinetic energy just after collision
  • converts into gravitational potential energy gain.
  1. Moment of inertia about the pivot

The rod is uniform, length l=1 ml=1\,\text{m}l=1m, mass M=2 kgM=2\,\text{kg}M=2kg, pivoted at one end.

So its moment of inertia about OOO is

Irod=13Ml2=13(2)(1)2=23.I_{\text{rod}}=\frac{1}{3}Ml^2=\frac{1}{3}(2)(1)^2=\frac{2}{3}.Irod​=31​Ml2=31​(2)(1)2=32​.

The block of mass m=1 kgm=1\,\text{kg}m=1kg sticks at the lower end of the rod, i.e. at distance l=1 ml=1\,\text{m}l=1m from the pivot. Hence

Iblock=ml2=(1)(1)2=1.I_{\text{block}}=ml^2=(1)(1)^2=1.Iblock​=ml2=(1)(1)2=1.

Therefore total moment of inertia after collision is

I=Irod+Iblock=23+1=53.I=I_{\text{rod}}+I_{\text{block}}=\frac{2}{3}+1=\frac{5}{3}.I=Irod​+Iblock​=32​+1=35​.
  1. Conservation of angular momentum during collision

Before collision, only the block has angular momentum about OOO.

Since the block hits the lower end perpendicularly,

Li=mvl=(1)(6)(1)=6.L_i = mvl = (1)(6)(1)=6.Li​=mvl=(1)(6)(1)=6.

After collision, if angular speed is ω\omegaω,

Lf=Iω=53ω.L_f = I\omega = \frac{5}{3}\omega.Lf​=Iω=35​ω.

Thus,

6=53ω6=\frac{5}{3}\omega6=35​ω ω=185=3.6 rad/s.\omega = \frac{18}{5}=3.6\,\text{rad/s}.ω=518​=3.6rad/s.
  1. Rotational kinetic energy just after collision
K=12Iω2=12⋅53⋅(3.6)2.K=\frac{1}{2}I\omega^2 =\frac{1}{2}\cdot \frac{5}{3}\cdot (3.6)^2.K=21​Iω2=21​⋅35​⋅(3.6)2.

Since

(3.6)2=12.96,(3.6)^2=12.96,(3.6)2=12.96,

we get

K=12⋅53⋅12.96=10.8 J.K=\frac{1}{2}\cdot \frac{5}{3}\cdot 12.96=10.8\,\text{J}.K=21​⋅35​⋅12.96=10.8J.
  1. Gain in potential energy at maximum angle θ\thetaθ

When the system rises by angle θ\thetaθ:

  • The rod’s center of mass rises by
l2(1−cos⁡θ).\frac{l}{2}(1-\cos\theta).2l​(1−cosθ).

So gain in PE of rod is

ΔUrod=Mgl2(1−cos⁡θ).\Delta U_{\text{rod}} = Mg\frac{l}{2}(1-\cos\theta).ΔUrod​=Mg2l​(1−cosθ).
  • The block at the lower end rises by
l(1−cos⁡θ).l(1-\cos\theta).l(1−cosθ).

So gain in PE of block is

ΔUblock=mgl(1−cos⁡θ).\Delta U_{\text{block}} = mgl(1-\cos\theta).ΔUblock​=mgl(1−cosθ).

Total gain:

ΔU=Mgl2(1−cos⁡θ)+mgl(1−cos⁡θ).\Delta U = Mg\frac{l}{2}(1-\cos\theta)+mgl(1-\cos\theta).ΔU=Mg2l​(1−cosθ)+mgl(1−cosθ).

Substitute M=2M=2M=2, m=1m=1m=1, g=10g=10g=10, l=1l=1l=1:

ΔU=2⋅10⋅12(1−cos⁡θ)+1⋅10⋅1(1−cos⁡θ)\Delta U = 2\cdot 10\cdot \frac{1}{2}(1-\cos\theta)+1\cdot 10\cdot 1(1-\cos\theta)ΔU=2⋅10⋅21​(1−cosθ)+1⋅10⋅1(1−cosθ) =10(1−cos⁡θ)+10(1−cos⁡θ)=10(1-\cos\theta)+10(1-\cos\theta)=10(1−cosθ)+10(1−cosθ) =20(1−cos⁡θ).=20(1-\cos\theta).=20(1−cosθ).

At the highest point,

K=ΔUK=\Delta UK=ΔU

so

10.8=20(1−cos⁡θ).10.8=20(1-\cos\theta).10.8=20(1−cosθ).

Hence

1−cos⁡θ=0.541-\cos\theta=0.541−cosθ=0.54 cos⁡θ=0.46.\cos\theta=0.46.cosθ=0.46.

Therefore,

θ=cos⁡−1(0.46)≈62.6∘.\theta = \cos^{-1}(0.46) \approx 62.6^\circ.θ=cos−1(0.46)≈62.6∘.

So approximately,

θ≈63∘\boxed{\theta\approx 63^\circ}θ≈63∘​
  1. Check options
  • A: 63∘63^\circ63∘ ✅
  • B: 69∘69^\circ69∘
  • C: 55∘55^\circ55∘
  • D: 49∘49^\circ49∘

Thus the correct option is A.

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