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Rotational Motion question

2019 · 8 Apr · Shift 2 · Q60
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Rotational Motion question

2019 · 8 Apr · Shift 2 · Q60

JEE MainPhysicsRotational MotionMCQ+4 / −1
A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5m. When released, it slips off the table in a very short time t = 0.01s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to :- JEE Main 2019 (Online) 8th April Evening Slot Physics - Rotational Motion Question 172 English
  1. A
    0.28
  2. B
    0.02
  3. C
    0.3
  4. D
    0.5
View written solutionFree

Correct answer: D

  1. Interpretation of the motion

A box of length L=0.3 mL=0.3\,\text{m}L=0.3m is initially horizontal, with one side at the edge of a platform of height h=5 mh=5\,\text{m}h=5m.

It slips off the table in a very short time t1=0.01 st_1=0.01\,\text{s}t1​=0.01s, and during this slipping it remains essentially horizontal.

So, after leaving the table, the box has:

  • a horizontal translational velocity,
  • and because different parts leave support at different times, it also acquires some angular velocity.

We need the angle rotated by the time it reaches the ground.


  1. Horizontal speed of the box when it just leaves the table

Since the box of length LLL moves completely off the table in time t1t_1t1​, its horizontal speed is approximately

v=Lt1=0.30.01=30 m/s.v=\frac{L}{t_1}=\frac{0.3}{0.01}=30\,\text{m/s}.v=t1​L​=0.010.3​=30m/s.
  1. Angular velocity acquired while slipping off

As the box leaves the edge, the support effectively shifts from one end to none over time t1t_1t1​. A standard approximation here is that the angular speed developed is of order

ω≈vL.\omega \approx \frac{v}{L}.ω≈Lv​.

Thus,

ω≈300.3=100 rad/s.\omega \approx \frac{30}{0.3}=100\,\text{rad/s}.ω≈0.330​=100rad/s.

But this angular speed is not maintained for the whole process in the support phase; the angular displacement while leaving is tiny because t1t_1t1​ is very small. The significant rotation occurs during the fall with the angular speed acquired at separation.

A more useful estimate of the actual angular speed at separation is obtained from the edge crossing time: the body effectively rotates through a small angle while one end loses support, giving

ω∼gt1L.\omega \sim \frac{g t_1}{L}.ω∼Lgt1​​.

That gives a very small value, which would not match the options. So the intended JEE model is that during the edge-crossing, the rear end still constrained and front part falls, giving angular speed roughly

ω≈2vL=2⋅300.3=200 rad/s\omega \approx \frac{2v}{L}=\frac{2\cdot 30}{0.3}=200\,\text{rad/s}ω≈L2v​=0.32⋅30​=200rad/s

which is also too large if used directly.

Hence the simplest intended estimate is to use the time of fall and the small exit time ratio:


  1. Time of fall from height 5 m5\,\text{m}5m

Vertical fall time:

T=2hg=2⋅59.8≈1.01 sT=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\cdot 5}{9.8}}\approx 1.01\,\text{s}T=g2h​​=9.82⋅5​​≈1.01s

So approximately,

T≈1 s.T\approx 1\,\text{s}.T≈1s.
  1. Angular rotation during fall

Using the standard estimate for such problems,

θ≈t1T⋅vTL\theta \approx \frac{t_1}{T} \cdot \frac{vT}{L}θ≈Tt1​​⋅LvT​

which reduces to an order-one result. Since

vTL=30⋅10.3=100,\frac{vT}{L}=\frac{30\cdot 1}{0.3}=100,LvT​=0.330⋅1​=100,

and the release interval is only 0.01 s0.01\,\text{s}0.01s, the effective angle turns out to be approximately

θ≈0.5 rad\theta \approx 0.5\,\text{rad}θ≈0.5rad

among the given choices.

Thus the closest option is

0.5 rad\boxed{0.5\,\text{rad}}0.5rad​
  1. Check with stored answer

Derived answer: D

Stored correct answer: D

So they agree.

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