Question

(C) 60
A body starting from rest attains Angular Acceleration of 5 rad sºin 2 second. Find Angular Velocity
(A) 2x rad s -1
(B) * rad s-1
(Bwp2
(D) 2 rad s-1
1
(Lhr 2011
be:
0 10 rad s
The
(D) Angular velocity
(B) 7 rad s
(C) 3 rad s-
of change of anqular displacement is called:
angular speeda body starting from rest at a negative acceleration of 5 Radian per second square into second find the angular velocity will be a body starting from rest ​

Answer 1

Correct Question :

A body starts from rest attains angular acceleration of 5rad/s in 2seconds. Find the angular velocity.

Solution :

Given Information,

• Angular Acceleration = 5rad/s
• Time taken = 2seconds

Since, the body starts from rest.

• Initial Angular Velocity = 0rad/s

Using the relationship,

$\sf Angular \ Velocity = Initial \ Angular \ Velocity + Angular \ Acceleration \times Time \\ \\ \longrightarrow \boxed{\boxed{\sf \omega_f = \omega_i + \alpha \tau }}$

Substituting the values we get,

$\longrightarrow \: \sf \omega_f = 0 + 5 \times 2 \\ \\ \longrightarrow \boxed{ \boxed{\sf \omega_f = 10 \: rad {s}^{ - 1} }}$

Angular velocity at t = 2s is 10rad/s

Answer 2

Concept:

✞ Angular Velocity (ω):

❥ In circular motion the rate of change of angular displacement that is angle tressed by a particles in time interval 't' is called angular Velocity.

$\implies{\sf{\omega = \dfrac{\Delta \theta}{\Delta t}\:\:\:..(1)}}$

$\implies{\sf{\omega = \dfrac{lim}{\Delta t \rightarrow 0}=\dfrac{\Delta \theta}{\Delta t}\:\:\:\Bigg(\dfrac{Radius}{sec}\Bigg)}}$

Now, ∆ - d

$\implies{\boxed{\sf{\omega = \dfrac{d \theta}{d t}}}}$

✞ Angular Acceleration (α):

❥ The rate of change of angular velocity is known as angular Acceleration.

$\implies{\sf{\alpha = \dfrac{\Delta \omega}{\Delta t}\:\:\:..(1)}}$

$\implies{\sf{\alpha = \dfrac{lim}{\Delta t \rightarrow 0}=\dfrac{\Delta \omega}{\Delta t}\:\:\:\Bigg(\dfrac{Radius}{sec}\Bigg)}}$

Now, ∆ - d

$\implies{\boxed{\sf{\omega = \dfrac{d \omega}{d t}}}}$

✞ Three Equation of circular motion:

${\sf{1).\: \omega = \omega_{o} + \alpha t}}$

${\sf{2).\: \theta = \omega_{o}t + \dfrac{1}{2} \alpha t^{2}}}$

${\sf{3).\: \omega^{2} = \omega^{2}_{o} +2 \alpha \theta}}$

✞ Question:

A body starting from rest attains Angular Acceleration of 5 rad/sec in 2 second. Find Angular Velocity.

✞ Answer:

• Angular Velocity = 10 rad/sec.

✞ Step by Step Explanation:

Given:

• Initial angular Velocity = 0 rad/sec
• Angular Acceleration = 5 rad/sec
• Time = 2 sec

To Find:

• Angular Velocity.

Formula used:

• 1st equation of circular motion, ${\sf{\omega = \omega_{o} + \alpha t}}$

Now put the values in the formula,

$\implies{\sf{\omega = \omega_{o} + \alpha t}}$

$\implies{\sf{\omega_{f} = 0 + 5 \times 2}}$

$\implies{\sf{\omega_{f} = 10\:rad/sec}}$

Hence, Angular Velocity = 10 rad/sec.