A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg-m2. If the string is pulled with a force F , the resulting angular acceleration of the pulley is 2 rad/s2. Determine the magnitude of the force F?

Respuesta :

Answer:

f = 8 N

Explanation:

from the question we are given the following values

radius of the pulley (r) = 0.05 m

moment of inertia (I) = 0.2 kg.m^{2}

angular acceleration (∝) = 2 rad/sec

force (f) = ?

  • Torque = force x radius = f x r
  • Torque = moment of inertia x angular acceleration = I x ∝
  • therefore torque = f x r = I x ∝

we can apply  f x r = I x ∝ to get our force

f x 0.05 = 0.2 x 2

0.05f = 0.4

f = 8 N