Two homogeneous rods are attached to the disk with a radius of 10cm,as in the figure, and the system can rotate freely araound the disk center...

Question

Two homogeneous rods are attached to the disk with a radius of 10cm,as in the figure, and the system can rotate freely araound the disk center. The mass of the disk is 800 grams and that of the rods is 300 grams. What will be the angular acceleration of the system when the forces F1 and F2 are applied as shown as figure?

Answer 1

It is given to us that

F₁ = 10 N

F₂ = 10 N

M_d = 0.8 kg

M_r = 0.3 kg

We need to find the angular acceleration of the system.

To solve the problem, we can use the Newton’s law for rotational motion

                                                            τ = α∙I .

Here

τ – torque,

α – angular acceleration,

I – total moment of inertia of the system.

Total moment of inertia of the system

                                                                        I = I₁ + I₂ + I₃

The moment of inertia of 2 rods are the same

                                               I₁ = I₂ = 1/12 (M_r∙ 4R²) =  1/3 ( M_r∙R²)

The moment of inertia of a disk

                                                                       I₃ = ½ M_d∙R²

Hence, total moment of inertia of the system

                                                      I = ⅔ ( M_r∙R²) + ½ M_d∙R² = R²(⅔M_r  + ½ M_d)

Torque

                                                                             τ = τ₁ – τ₂

Or

                                                        τ = F₁R – F₂ R /√2 = R (F₁ – F₂/√2)

Now we can write

                                                       R (F₁ – F₂/√2) = α∙ R²(⅔M_r  + ½ M_d)

Finally,

                                                          α = (F₁ – F₂/√2) / R(⅔M_r  + ½ M_d)

After plugging in the values, we have:

                                     α = (10 N – 10 N/√2) / 0.1 m (⅔ 0.3 kg + ½ 0.8 kg)

 Angular acceleration of the system

α = 50 rad/s²