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A ball is spun with angular acceleration $\mathrm{\alpha=6 \mathrm{t}^{2}-2 \mathrm{t}}$ where t is in second and $\mathrm{\alpha}$ is in rad s^-2. $\text { At } \mathrm{t}=0 \text {, }$ the ball has angular velocity of $\mathrm{10 \; \mathrm{rads}^{-1}}$ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :
Option: 1
$\mathrm{\frac{3}{2} t^{4}-t^{2}+10 t}$

Option: 2
$\mathrm{\frac{t^{4}}{2}-\frac{t^{3}}{3}+10 t+4}$

Option: 3
$\mathrm{\frac{2 t^{4}}{3}-\frac{t^{3}}{6}+10 t+12}$

Option: 4
$\mathrm{2 t^{4}-\frac{t^{3}}{2}+5 t+4}$

Answers (1)

best_answer

$\mathrm{\alpha=6 t^{2}-2 t }$

$\mathrm{\frac{d \omega}{d t}=6 t^{2}-2 t }$

$\mathrm{\int_{10}^{\omega} d \omega=\int_{t=0}^{t}\left(6 t^{2}-2 t\right) d t \\ }$

$\mathrm{\omega-10=\left [\frac{6t^{3}}{3}\frac{-2t^{2}}{2} \right ]_{0}^{t} }$

$\mathrm{\omega=10+2 t^{3}-t^{2} }$ --------(1)

$\mathrm{\frac{d\theta }{dt}=10+2t^{3}-t^{2}}$

$\mathrm{\int_{\theta=4}^{\theta} d \theta=\int_{t=0}^{t}\left(10+2 t^{3}-t^{2}\right) d t }$

$\mathrm{\theta-4=\left[10 t+\frac{2 t^{4}}{4}-\frac{t^{3}}{3}\right]_{0}^{t} \\ }$

$\mathrm{\theta=\frac{t^{4}}{2}-\frac{t^{3}}{3}+10 t+4 }$

Hence 2 is correct answer.

Posted by

himanshu.meshram

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