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

Answer:

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

Explanation:

From the question we are told that

              The mass is  $m = 50kg$

              The diameter is  $d = 0.12m$

              The radius is $r = \frac{d}{2} = \frac{0.12}{2} =0.06m$

              The force is $F = 9.0N$

               The distance is  $s = 2.0m$

The force is mathematically represented as

                    $F = ma$

                    $9 = 50 * a$

                   $a = \frac{9}{50}$

                      $= 0.18m/s^2$

From the Equation of motion we have

           $v^2 = u^2 +2as$

$v$ is the final speed of the cable

substituting values we have

          $v^2 = 0 +2 (0.18)(2)$

              $= 0.72 m^2 /s^2$

          $v = \sqrt{0.72}$

            $= 0.84 m/s$

Now the linear velocity can be mathematically represented as

                  $v = wr$

now substituting values

                 $0.84 = w * 0.06$

=>        $w = \frac{0.84}{0.06}$

               $= 14 rad/s$

Answer 2

Answer:

the final velocity of the cable is 0.84m/s

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

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