Question

b. 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.

Answer 1

f = 8 N

Step-by-step explanation:

Data provided in the question

Radius of the pulley = r = 0.05 m

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

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

Torque is

`= force * radius`

`= f * r`

And,

Torque is also

`= moment\ of\ inertia * angular\ acceleration`

`= I * \alpha`

So,

We can say that

`f * r = I * \alpha`

`f * 0.05 = 0.2 * 2`

0.05f = 0.4

f = 8 N

We simply applied the above formulas