Solve the following :


A light rod of length $1 \mathrm{~m}$ is pivoted at its centre and two masses of $5 \mathrm{~kg}$ ang $2 \mathrm{~kg}$ are hung from the ends as shown in figure. Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning.


$\tau_{\text {Net }}=50 \times 0.5-20 \times 0.5$

$\tau_{\text {Net }}=15 \mathrm{~N}-\mathrm{m}$

$\tau=I \alpha$

$15=\left[(2)(0.5)^{2}+5(0.5)^{2}\right] \alpha$

$\alpha=8.57 \sec ^{2}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now