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..

## Comments

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TopNewestThe 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 · 3 years, 9 months ago

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– Joe Bobby · 1 year, 7 months ago

But L=1m is not given right??Log in to reply

– Max B · 3 years ago

nice......just mingle around the formulas...as simple as that.Log in to reply

– Krishna Jha · 3 years, 9 months ago

Thanks...Log in to reply