A cylinder rotating at an angular speed of 50 revolution/sec is brough

1.	A cylinder rotating at an angular speed of 50 revolution/sec is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed. ?
Answer» Given in the question :- Assume , Time = t second. $\omega$ Init. velocity = 50 revolution / second. $\omega'$ Fin. velocity = $\alpha = 1 revoution/second^2$ In case of first cyl. $\omega' = \omega - \alpha t$ ω' = 50 - 1 × t ω' = 50 - t In case of second cyl. Initial speed = 0 and FINAL speed = ω' $\omega' = \omega - \alpha t$ put the value of ω' 50 - t = t × 1 *t = 25 second.* *Hope it Helps :-)*

A cylinder rotating at an angular speed of 50 revolution/sec is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed. ?

Answer»

Given in the question :-

Assume , Time = t second.

$\omega$ Init. velocity = 50 revolution / second.

$\omega'$ Fin. velocity =

$\alpha = 1 revoution/second^2$

In case of first cyl.

$\omega' = \omega - \alpha t$

ω' = 50 - 1 × t

ω' = 50 - t

In case of second cyl.

Initial speed = 0 and FINAL speed = ω'

$\omega' = \omega - \alpha t$

put the value of ω'

50 - t = t × 1

t = 25 second.

Hope it Helps :-)

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