Question

A light cable is wrapped around a solid cylinder of mass 50kg and diameter 0.12m. The cylinder rotates about a stationary axis. The cable is pulled with a constant force of 9.0N for a distance of 2.0m. The cable unwinds without slipping, turning the cylinder as it does. The cylinder is initially at rest.Find it angular velocity w and the final speed v of the cable.

Answer 1

the final velocity of the cable is 0.84m/s

the angular velocity is of the cable is 14rad/s

Step-by-step explanation:

from the equation of motion

`v^2 = u^2 + 2as`

F = ma

9.0N = 50kg × a

a = 0.18m/s²

`v^2 = u^2 + 2as`

v² = 0 + 2(0.18)(2)

v² = 0.72m²/s²

v = √0.72m²/s²

v = 0.84m/s

Hence, the final velocity of the cable is 0.84m/s

b)

v = rω

ω = v/r

r = d/2 = 0.12m/2

= 0.06m

ω = 0.84 / 0.06

= 14rad/s

Hence , the angular velocity is of the cable is 14rad/s

Answer 2

The angular velocity is
`w = 14\ rad /s` and the final velocity is v = 0.84 m/s

Step-by-step explanation:

From the question we are told that

The mass is
`m = 50kg`

The diameter is
`d = 0.12m`

The radius is
`r = (d)/(2) = (0.12)/(2) =0.06m`

The force is
`F = 9.0N`

The distance is
`s = 2.0m`

The force is mathematically represented as

`F = ma`

`9 = 50 * a`

`a = (9)/(50)`

`= 0.18m/s^2`

From the Equation of motion we have

`v^2 = u^2 +2as`

`v` is the final speed of the cable

substituting values we have

`v^2 = 0 +2 (0.18)(2)`

`= 0.72 m^2 /s^2`

`v = √(0.72)`

`= 0.84 m/s`

Now the linear velocity can be mathematically represented as

`v = wr`

now substituting values

`0.84 = w * 0.06`

=>
`w = (0.84)/(0.06)`

`= 14 rad/s`