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Rotational Motion question

2016 · Q149
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Rotational Motion question

2016 · Q149

NEETPhysicsRotational MotionMCQ+4 / −1
A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s−-−2. Its net acceleration in m s−-−2 at the end of 2.0 s is approximately
  1. A
    6.0
  2. B
    3.0
  3. C
    8.0
  4. D
    7.0
View written solutionFree

Correct answer: C

Given, r = 50 cm = 0.5 m, α\alpha α = 2.0 rad s–2 , ω0{\omega _0}ω0​ = 0

At the end of 2 s,
Tangential acceleration, at
= rα\alpha α = 0.5 × 2 = 1 m s–2

Radial acceleration, = ar=ω2r=(ω0+αt)2r{a_r} = {\omega ^2}r = {\left( {{\omega _0} + \alpha t} \right)^2}rar​=ω2r=(ω0​+αt)2r

= (0 + 2 × 2)2 × 0.5 = 8 m s–2

∴\therefore∴ Net acceleration,
a=at2+ar2=12+82=65≈8 ms−2a = \sqrt {a_t^2 + a_r^2} = \sqrt {{1^2} + {8^2}} = \sqrt {65} \approx 8\,m{s^{ - 2}}a=at2​+ar2​​=12+82​=65​≈8ms−2

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