Question

Plz solve it..................i will mark him as brainlist​

Answer 1

Explanation:

Displacement θ=

20

t

2

+

5

t

Angular velocity,

ω=

dt

dθ

=

20

2t

+

5

1

=

10

t

+

5

1

dt

dθ

∣

t=4

=

10

4

+

5

1

=0.4+0.2=0.6

∴k=0.6

∴5k=0.6×5=3

Answer 2

Given :

• A particle starts from rest

• Its angular displacement is given by $\sf{\theta=\dfrac{t^2}{20}+\dfrac{t}{5}}$

To find :

• Angular velocity of body at the end of t = 4 s; ω = ?

Knowledge required :

• Formula to calculate angular velocity

Angular velocity is known as the rate of change of angular displacement, with respect to time.

It is calculated by the formula

$\red{\bigstar}\boxed{\sf{\omega=\dfrac{d\; \theta}{d\; t}}}$

[ Where ω is angular velocity, (d θ) is change in angular displacement and (d t) is change in time ]

Solution :

Using formula

$\longrightarrow \sf{\omega=\dfrac{d\;\theta}{d\;t}}$

$\longrightarrow \sf{\omega(t)=\dfrac{d\;\bigg(\dfrac{t^2}{20}+\dfrac{t}{5}\bigg)}{d\;t}}$

Differentiating LHS

$\longrightarrow \sf{\omega(t)=2\times\dfrac{t}{20}+1\times\dfrac{1}{5}}$

$\longrightarrow \sf{\omega(t)=\dfrac{t}{10}+\dfrac{1}{5}}$

Now calculating angular velocity of particle at the end of t = 4 sec

$\longrightarrow \sf{\omega(4)=\dfrac{(4)}{10}+\dfrac{1}{5}}$

$\longrightarrow \sf{\omega(4)=\dfrac{4+2}{10}}$

$\longrightarrow\underline{\underline{\red{ \sf{\omega(4)=\dfrac{6}{10}=\dfrac{3}{5}\;\;Rad\;sec^{-1}}}}}$

Hence,

• Angular velocity of particle at the end of t = 4 s will be $\bf{\dfrac{3}{5}\;\;Rad\;s^{-1}}$.