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Rotational Mechanics Question

Q. A uniform metre stick of mass M is hinged at one end and supported in the horizontal direction by a string attached to the other end. What should be the initial angular acceleration of the stick if the string is cut?

a. \(\frac{3}{2}g\hspace{2 mm}rad\hspace{1 mm}s^{-2}\)

b. \(g\hspace{2 mm}rad\hspace{1 mm}s^{-2}\)

c. \(3g\hspace{2 mm}rad\hspace{1 mm}s^{-2}\)

d. \(4g\hspace{2 mm}rad\hspace{1 mm}s^{-2}\)

The answer given was a... But i think it should be c. Pls solve this..

Note by Krishna Jha
4 years, 1 month ago

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The stick has moment of inertia \(I = \tfrac{1}{3}ML^2\) about the hinge. The weight of the stick is \(W = Mg\) and it acts at the center of mass which is \(d = \tfrac{1}{2}L\) from the hinge. Initially, the weight is perpendicular to the stick. So, the total torque initially is \(\tau = Wd = \tfrac{1}{2}MgL\). Since \(\tau = I\alpha\), the initial angular acceleration is \(\alpha = \dfrac{\tau}{I} = \dfrac{\tfrac{1}{2}MgL}{\tfrac{1}{3}ML^2} = \dfrac{3g}{2L}\). Plug in \(L = 1m\) to get the answer. Jimmy Kariznov · 4 years, 1 month ago

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@Jimmy Kariznov But L=1m is not given right?? Joe Bobby · 1 year, 11 months ago

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@Jimmy Kariznov nice......just mingle around the formulas...as simple as that. Max B · 3 years, 4 months ago

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@Jimmy Kariznov Thanks... Krishna Jha · 4 years, 1 month ago

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