First slide

Rotational motion

Question

One of the strings at end Q is cut at t = 0 and the rod starts rotating about the other end P. Then

Difficult

A

At t = 0, angular acceleration of rod about P is $\frac{3 g}{2 L} .$
B

At t = 0, angular acceleration of rod about C.M. of rod is $\frac{3 g}{2 L}$ .
C

At t = 0, acceleration of C.M. of rod is $\frac{3 g}{4}$ in downward direction.
D

At t = 0, tension in the string at P is $\frac{Mg}{4}$

Solution

$\begin{array}{l} Mg (\frac{L}{2}) = (\frac{{ML}^{2}}{3}) α \\ ∴ α = \frac{3 g}{2 L} \\ ∴ a_{c o m} = α (\frac{L}{2}) = \frac{3 g}{2 L} (\frac{L}{2}) = \frac{3 g}{4} \\ - T + Mg = M (\frac{3 g}{4}) \\ ∴ T = \frac{Mg}{4} \end{array}$

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