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.

The angular velocity is and the final velocity is  v = 0.84 m/s

Explanation:

From the question we are told that

              The mass is  

              The diameter is  

              The radius is

              The force is

               The distance is  

The force is mathematically represented as

From the Equation of motion we have

is the final speed of the cable

substituting values we have

Now the linear velocity can be mathematically represented as

now substituting values

=>        

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

F = ma

9.0N = 50kg × a

a = 0.18m/s²

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

lol

explanation:

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.