Physics, asked by shneerthi, 5 months ago

the particle is revolving uniformly at 170 rad s-¹ is brought to the rest by a uniform angular retardation of
 \alpha  =
5 rad s-²


find the time required to stop​

Answers

Answered by Cynefin
101

 \LARGE{ \underline{ \blue{ \sf{Required \: answer:}}}}

The particle is in circular motion and we have the different quantities like Initial & Final angular velocity, angular acceleration.

  • Initial angular velocity = 170 rad/s
  • Final angular velocity = 0 m/s
  • Angular acceleration = -5 rad/s²

We have to find the time required to stop the p.

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By using 1st equation of motion,

 \large{ \because{ \underline{ \boxed{ \rm{ \omega =  \omega_0+  \alpha t}}}}}

Putting the values of w, w0 and \alpha:

 \rm{0 = 170 + ( - 5)t}

 \rm{5t = 170}

Dividing 5 from both sides to get t

 \rm{t = 170 \div 5 = 34 \: s}

Thus the time taken by the particle to stop with the given quantities is 34 s. And we are done...


amitkumar44481: Perfect :-)
Anonymous: Nice ♥️
Cynefin: Thank uh :)
Draxillus: Awesome :-)
Answered by Qᴜɪɴɴ
74

Given:

  • Initial angular velocity = ω°= 170rad/ sec
  • Final angular velocity = ω = 0 rad/sec
  • Angular acceleration =
  •   \alpha  =  - 5m /{s}^{2}

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Need to find:

  • What's the time it needs to stop ?

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Solution:

We know,

 { \omega}^{2}  = { \omega°}  +   \alpha  t

Now putting values we get

0 = 170 + ( - 5)t

 \implies \:  - 5t =  - 170

 \implies \: t = 34sec

Thus the time required is \red{\bold{\boxed{\large{34sec}}}}


Cynefin: Awesome
Anonymous: Great:)
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