Question

A body whose moment of inertia is 3 kg metre square is at rest it is rotating for 20 seconds with a moment of force 6 newton metre find the angular displacement of body

Answer 1

Answer:

• Angular Displacement (θ) = 400 Radians.

Given:

1. Moment of Inertia (I) = 3 Kg/m².
2. Time Period (T) = 20 Sec.
3. Torque (τ) = 6 N-m.

Explanation:

$\rule{300}{1.5}$

From the Relation we Know,

$\large\bigstar \;{\boxed{\tt \tau = I \alpha}}$

$\bold{Here}\begin{cases}\tau \; \text{Denotes Torque} \\ \text{ I Denotes Moment of Inertia} \\ \alpha \; \text{Denotes angular Acceleration}\end{cases}$

Now,

$\large{\boxed{\tt \tau = I \alpha}}$

Substituting The Values.

$\large{\tt \hookrightarrow 6 \; Nm = 3 \; Kg/m^2 \times \alpha}$

$\large{\tt \hookrightarrow 6 = 3 \times \alpha}$

$\large{\tt \hookrightarrow \alpha = \dfrac{6}{3}}$

$\large{\tt \hookrightarrow \alpha = \cancel{\dfrac{6}{3}}}$

$\large{\underline{\boxed{\tt \alpha = 2 \; Rad/sec^2}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, From Angular Kinematic equation

$\large\bigstar \;{\boxed{\tt \theta = \omega t + \dfrac{1}{2}\alpha t^2}}$

$\bold{Here}\begin{cases}\omega \; \text{Denotes Angular Velocity} \\ \text{ t Denotes Time taken} \\ \alpha \; \text{Denotes angular Acceleration} \\ \theta \; \text{Denotes Angular Displacement}\end{cases}$

Now,

$\large{\boxed{\tt \theta = \omega t + \dfrac{1}{2}\alpha t^2}}$

Substituting The Values.

$\large{\tt \hookrightarrow \theta = 0 \times t + \dfrac{1}{2} \times 2 \times (20)^2}$

• Initial Angular Velocity is Zero.

$\large{\tt \hookrightarrow \theta = 0 + \dfrac{1}{2} \times 2 \times 400}$

$\large{\tt \hookrightarrow \theta = \dfrac{2}{2} \times 400}$

$\large{\tt \hookrightarrow \theta = \cancel{\dfrac{2}{2}} \times 400}$

$\large{\tt \hookrightarrow \theta = 1 \times 400}$

$\huge{\boxed{\boxed{\tt \theta = 400 \; Rad}}}$

∴ Angular Displacement (θ) is  400 Radians.

$\rule{300}{1.5}$