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Alinara [238K]
3 years ago
11

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.
Physics
1 answer:
Oduvanchick [21]3 years ago
8 0

Answer:

f = 8 N

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 \times  radius

= f \times r

And,

Torque is also

= moment\ of\ inertia \times angular\ acceleration

= I \times \alpha

So,

We can say that

f \times r = I \times \alpha

f \times 0.05 = 0.2 \times 2

0.05f = 0.4

f = 8 N

We simply applied the above formulas

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