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An automobile is started from rest with one of its doors initially at right angles. lf the hinges of the door are toward the front of the car, the door will shut as the automobile picks up speed. The acceleration a is constant and the centre of mass is at a distance d from the hinges. The time T needed for the doors to close is given by:
[Given:π20dθsinθ=N](Consider the door as a square plate)

  1. T=(3d2a)N
  2. T=(d3a)N
  3. T=(2da)N
  4. T=(2d3a)N

A
T=(d3a)N
B
T=(2da)N
C
T=(3d2a)N
D
T=(2d3a)N
Solution
Verified by Toppr

Equation of motion is
Iα=Ma(dcosθ)
Now moment of inertia of the door about the hinges is given as
I=Md23+Md2
=M(43d2)
Hence,
43Md2α=Madcosθ

43d×ωdωdθ=acosθ(α=ωdωdθ)

43dωdω=acosθdθ

43dω22=asinθ

ω2=3a2dsinθ+C

where, C=constant of integration

At θ=0 (initially), we have θ=0
C=0
ω2=3a2dsinθ

ω=3a2dsinθ

dθsinθ=3a2ddt and integrating between t=0 and t=T
(corresponding to θ=0 and θ=π2)

π20dθsinθ=3a2d.T

T=(2d3a)N

since,given:N=π20dθsinθ

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