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Topic: Equilibrium of rigid bodies, rigid body rotation, equations of rotational motion· Rotational Motion

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 acc…

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 \(ms^{-2}\) at the end of 2.0 s is approximately :

1. 8.0

2. 7.0

3. 6.0

4. 3.0

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Answer: 1. 8.0

Key idea: net acceleration combines tangential and centripetal parts.

  1. \(\omega=\alpha t=2\times2=4\) rad/s.
  2. \(a_t=\alpha r=2\times0.5=1\) m/s\(^2\); \(a_c=\omega^2r=16\times0.5=8\) m/s\(^2\).
  3. \(a=\sqrt{1^2+8^2}=\sqrt{65}\approx8.06\approx8.0\) m/s\(^2\).
Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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