A cylinder of mass $5kg$ and radius $30cm$ and free to rotate about its axis, receives an angular impulse of $3kgm^2 s^{-1}$ followed by a similar impulse after every $4$ seconds. What is the angular speed of the cylinder $30s$ after the initial impulse? The cylinder is at rest initially.
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$m = 5kg, r = 30 \times 10^{-2}m, dL = 3kgm^2s^{-1}, t = 4 sec$.
Angular Impulse = dL = 3
We know, $\text{L}=\text{I}\omega$
$\therefore\text{dL}=3=\text{I}(\omega_{\text{f}}-\omega_{\text{i}})=\frac{1}{2}\text{mr}^2(\omega_{\text{f}-\omega_{\text{i}}})$
$3=\frac{1}{2}\times5\times(0.30)^2\times(\omega-0)$
Solving, $\omega=\frac{40}{3}\text{ rads}^{-1}$
Using $\omega=\omega_{\text{i}}+\alpha\text{t},\text{ we get}$
$\frac{40}{3}=0+\alpha(4)(\text{i.e.,})\alpha=\frac{10}{3}\text{ rads}^{-1}$
Since time gap between impulse acting is 4s, the impulse will go up to 32 s
$\therefore\omega=\omega_{\text{i}}+\alpha\text{t}=0\frac{10}{3}\times32=106.67\text{ rads}^{-1}$
Angular Impulse = dL = 3
We know, $\text{L}=\text{I}\omega$
$\therefore\text{dL}=3=\text{I}(\omega_{\text{f}}-\omega_{\text{i}})=\frac{1}{2}\text{mr}^2(\omega_{\text{f}-\omega_{\text{i}}})$
$3=\frac{1}{2}\times5\times(0.30)^2\times(\omega-0)$
Solving, $\omega=\frac{40}{3}\text{ rads}^{-1}$
Using $\omega=\omega_{\text{i}}+\alpha\text{t},\text{ we get}$
$\frac{40}{3}=0+\alpha(4)(\text{i.e.,})\alpha=\frac{10}{3}\text{ rads}^{-1}$
Since time gap between impulse acting is 4s, the impulse will go up to 32 s
$\therefore\omega=\omega_{\text{i}}+\alpha\text{t}=0\frac{10}{3}\times32=106.67\text{ rads}^{-1}$
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