Solve the following :
Question:

A dumb-bell consists of two identical small balls of mass $1 / 2 \mathrm{~kg}$ each connected to the two ends of a $50 \mathrm{~cm}$ long light rod. The dumb-bell is rotating about a fixed axis through the centre of the rod and perpendicular to it at an angular speed of $10 \mathrm{rad} / \mathrm{s}$. An impulsive force of average magnitude $5.0 \mathrm{~N}$ acts on one of the masses in the direction of its velocity for $0.10 \mathrm{~s}$. Find the new angular velocity of the system.

Solution:

$\mathrm{T}=\mathrm{I \omega}$

$\mathrm{F} \times \mathrm{r}=\left(m r^{2}+m r^{2}\right) \alpha$

$5(0.25)=2\left(\frac{1}{2}\right)(0.5)^{2} \alpha$

$\alpha=20 \mathrm{rad} / \mathrm{s}^{2}$

$\omega_{0}=10 \mathrm{rad} / \mathrm{sec}$

$\mathrm{T}=0.1$

$\omega=\omega_{0}+\alpha_{\mathrm{T}}$

$\omega=12 \mathrm{rad} / \mathrm{sec}$

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