Question

If the angular velocity of a merry-go-round is 60 degree
sec and you are 3.5 m from the centre of rotation
your linear velocity will be :-
(1)
22
m/s
7.
(2)
7
- m/s
3
7 п
(3) m/s
6
(4) at m/s

Answer 1

Answer:

answer is 3 ,7π/6 m/s

Explanation:

it is the most correct answer and I am damm sure .

Answer 2

Answer:

Explanation:

• $w=60 degree$$/ sec$,
• $1 degree= \pi /180$  
• angular velocity is find by,

                     ∴$w= 60$ ×$\pi$/180

                       $w=\pi /3$  $rad/sec$

• given ' r ' =  3.5m

         formula for linear velocity is given as ,

                 $v =rw$

•  lets substitute all teh values,

                       $v=\pi /3$ ×  $3.5$

                          $= 22/7$ ×  $1/3$ ×  $3.5$   (∴ $\pi =22/7$ )

                          $=11/2$ $m/s$   or

              'multiply  both up and down with 2',

                          $=22/6$

           'now divide both up and down with 7'

                          $=7\pi /6$ - $m/s$.

∴ the linear velocity will be 7$\pi$ / 6 or 11/3 .