Question

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​

Answer 1

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

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.

━━━━━━━━━━━━━━━━━━━━

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...

Answer 2

Given:

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

━━━━━━━━━━━━━━

Need to find:

• What's the time it needs to stop ?

━━━━━━━━━━━━━━━━━

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 $\red{\bold{\boxed{\large{34sec}}}}$