Question

If angular displacemebt of object is given by theta=7t+ t^2 + t^3 where time is seconds.its angular accleration at 1 sec is

Answer 1

Explanation:

differentiation of angular velocity is numarically angular acceleration...

Answer 2

Answer:

The angular acceleration at t = 1 s is 8 rad/s².

Explanation:

We have given that,

Angular displacement $\displaystyle{\sf\:(\:\theta\:)\:=\:7t\:+\:t^2\:+\:t^3}$

Time ( t ) is in seconds.

We have to find the angular acceleration $\displaystyle{\sf\:(\:\alpha\:)}$ at t = 1 s.

We know that, angular velocity is given by

$\displaystyle{\boxed{\pink{\sf\:\omega\:=\:\dfrac{d\:\theta}{dt}\:}}}$

$\displaystyle{\implies\sf\:\omega\:=\:\dfrac{d}{dt}\:(\:7t\:+\:t^2\:+\:t^3\:)}$

$\displaystyle{\implies\sf\:\omega\:=\:\dfrac{d}{dt}\:(\:7t\:)\:+\:\dfrac{d}{dt}\:(\:t^2\:)\:+\:\dfrac{d}{dt}\:(\:t^3\:)}$

$\displaystyle{\implies\sf\:\omega\:=\:7\:\times\:1\:+\:2t^{2\:-\:1}\:+\:3t^{3\:-\:1}}$

$\displaystyle{\implies\sf\:\omega\:=\:7\:+\:2t\:+\:3t^2}$

$\displaystyle{\implies\:\boxed{\green{\sf\:\omega\:=\:3t^2\:+\:2t\:+\:7\:}}}$

Now, we know that,

$\displaystyle{\boxed{\blue{\sf\:\alpha\:=\:\dfrac{d\:\omega}{dt}\:}}}$

$\displaystyle{\implies\sf\:\alpha\:=\:\dfrac{d}{dt}\:(\:3t^2\:+\:2t\:+\:7\:)}$

$\displaystyle{\implies\sf\:\alpha\:=\:\dfrac{d}{dt}\:(\:3t^2\:)\:+\:\dfrac{d}{dt}\:(\:2t\:)\:+\:\dfrac{d}{dt}\:(\:7\:)}$

$\displaystyle{\implies\sf\:\alpha\:=\:3\:\times\:2\:t^{2\:-\:1}\:+\:2\:\times\:1\:+\:0}$

$\displaystyle{\implies\sf\:\alpha\:=\:6t\:+\:2}$

Now, t = 1 s,

$\displaystyle{\therefore\:\sf\:\alpha\:=\:6\:(\:1\:)\:+\:2}$

$\displaystyle{\implies\sf\:\alpha\:=\:6\:+\:2}$

$\displaystyle{\implies\:\underline{\boxed{\red{\sf\:\alpha\:=\:8\:rad\:/\:s^2\:}}}}$

∴ The angular acceleration at t = 1 s is 8 rad/s².