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Rotational Motion question

2024 · 9 Apr · Shift 1 · Q88
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Rotational Motion question

2024 · 9 Apr · Shift 1 · Q88

JEE MainPhysicsRotational MotionNumerical+4 / −1
A string is wrapped around the rim of a wheel of moment of inertia 0.40 kgm20.40 \mathrm{~kgm}^20.40 kgm2 and radius 10 cm10 \mathrm{~cm}10 cm. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of 40 N40 \mathrm{~N}40 N. The angular velocity of the wheel after 10 s10 \mathrm{~s}10 s is x rad/sx \mathrm{~rad} / \mathrm{s}x rad/s, where xxx is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 100

  1. Given data
  • Moment of inertia: I=0.40 kg m2I = 0.40\,\mathrm{kg\,m^2}I=0.40kgm2
  • Radius of wheel: r=10 cm=0.10 mr = 10\,\mathrm{cm} = 0.10\,\mathrm{m}r=10cm=0.10m
  • Pulling force: F=40 NF = 40\,\mathrm{N}F=40N
  • Initial angular velocity: ω0=0\omega_0 = 0ω0​=0
  • Time: t=10 st = 10\,\mathrm{s}t=10s
  1. Find the torque on the wheel

The force is applied tangentially through the string, so torque is

τ=Fr\tau = F rτ=Fr

τ=40×0.10=4 N m\tau = 40 \times 0.10 = 4\,\mathrm{N\,m}τ=40×0.10=4Nm

  1. Find angular acceleration

Using rotational form of Newton's second law,

τ=Iα\tau = I\alphaτ=Iα

So,

α=τI=40.40=10 rad/s2\alpha = \frac{\tau}{I} = \frac{4}{0.40} = 10\,\mathrm{rad/s^2}α=Iτ​=0.404​=10rad/s2

  1. Find angular velocity after 10 s

Using

ω=ω0+αt\omega = \omega_0 + \alpha tω=ω0​+αt

ω=0+10×10=100 rad/s\omega = 0 + 10 \times 10 = 100\,\mathrm{rad/s}ω=0+10×10=100rad/s

  1. Final answer

x=100x = 100x=100

The derived answer matches the stored correct answer.

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