Question

The angular velocity of the particle is given by = 2
^3 − 6
^2 + 3.

Find the time when its angular acceleration becomes zero. ​

Answer 1

Correct Question:

The angular velocity of the particle is given by = 2  t³ − 6  t² + 3.  Find the time when its angular acceleration becomes zero. ​

Answer:

• Time (t) period is 2 seconds

Given:

1. ω = 2 t³ - 6 t² + 3

Explanation:

$\rule{300}{1.5}$

We know,

⇒ ω = 2  t³ − 6  t² + 3

Differentiating w.r.t time.

$\displaystyle\longrightarrow\sf \dfrac{d\;\omega}{dt}=\dfrac{d\;(2t^{3}-6t^{2}+3)}{dt}\\\\\\\longrightarrow\sf \alpha =\dfrac{d\;(2t^{3}-6t^{2}+3)}{dt} \ \ \ \because\Bigg[\dfrac{d\;\omega}{dt}=\alpha \Bigg]\\\\\\\longrightarrow\sf \alpha =6t^{\;(3-1)}-12t^{\;(2-1)}+0\\\\\\\longrightarrow\sf \alpha = 6t^{2}-12t\\\\\\\longrightarrow \underline{\boxed{\sf \alpha = 6t^{2}-12t\;\;rad/sec^{2}}}$

∴ We got angular acceleration.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now we need to find time when angular acceleration is zero, i.e. $\sf \alpha = 0$

Substituting and solving,

$\displaystyle\longrightarrow\sf \alpha = 6t^{2}-12t\\\\\\\longrightarrow\sf 0 = 6t^{2}-12t\\\\\\\longrightarrow\sf 6t^{2}=12t\\\\\\\longrightarrow\sf 6t=12\\\\\\\longrightarrow\sf t=\dfrac{12}{6}\\\\\\\longrightarrow\sf t = 2\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf t=2\;sec}}}}$

∴ At t = 2 s angular acceleration will be zero.

$\rule{300}{1.5}$

Answer 2

Answer:

Correct Question:

The angular velocity of the particle is given by = 2 t³ − 6 t² + 3. Find the time when its angular acceleration becomes zero.

Answer:

Time (t) period is 2 seconds

Given:

ω = 2 t³ - 6 t² + 3

Explanation:

\rule{300}{1.5}

We know,

⇒ ω = 2 t³ − 6 t² + 3

Differentiating w.r.t time.

\begin{gathered}\displaystyle\longrightarrow\sf \dfrac{d\;\omega}{dt}=\dfrac{d\;(2t^{3}-6t^{2}+3)}{dt}\\\\\\\longrightarrow\sf \alpha =\dfrac{d\;(2t^{3}-6t^{2}+3)}{dt} \ \ \ \because\Bigg[\dfrac{d\;\omega}{dt}=\alpha \Bigg]\\\\\\\longrightarrow\sf \alpha =6t^{\;(3-1)}-12t^{\;(2-1)}+0\\\\\\\longrightarrow\sf \alpha = 6t^{2}-12t\\\\\\\longrightarrow \underline{\boxed{\sf \alpha = 6t^{2}-12t\;\;rad/sec^{2}}}\end{gathered}

⟶

dt

dω

=

dt

d(2t

3

−6t

2

+3)

⟶α=

dt

d(2t

3

−6t

2

+3)

∵[

dt

dω

=α]

⟶α=6t

(3−1)

−12t

(2−1)

+0

⟶α=6t

2

−12t

⟶

α=6t

2

−12trad/sec

2

∴ We got angular acceleration.

\rule{300}{1.5}

\rule{300}{1.5}

Now we need to find time when angular acceleration is zero, i.e. \sf \alpha = 0α=0

Substituting and solving,

\begin{gathered}\displaystyle\longrightarrow\sf \alpha = 6t^{2}-12t\\\\\\\longrightarrow\sf 0 = 6t^{2}-12t\\\\\\\longrightarrow\sf 6t^{2}=12t\\\\\\\longrightarrow\sf 6t=12\\\\\\\longrightarrow\sf t=\dfrac{12}{6}\\\\\\\longrightarrow\sf t = 2\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf t=2\;sec}}}}\end{gathered}

⟶α=6t

2

−12t

⟶0=6t

2

−12t

⟶6t

2

=12t

⟶6t=12

⟶t=

6

12

⟶t=2

⟶

t=2sec

∴ At t = 2 s angular acceleration will be zero.

\rule{300}{1.5}