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A cylinder having a mass of 2.0 kg can rotate about itscentral
axis through point O. Forces are applied: F1 = 6.0N; F2 =4.0N; F3 =
2.0 N and F4 = 5.0N. Also, R1=5.0 cm and R2 = 12 cm.Find the
magnitude and direction of the angular acceleration of thecylinder.
(During the rotation, the forces maintain their sameangles relative
to the cylinder).

Respuesta :

Answer:

The angular acceleration of the cylinder is 9.72 rad/s^2.

Explanation:

Given that,

Mass of cylinder = 2.0 kg

Force F₁=6.0 N

Force F₂=4.0 N

Force F₃=2.0 N

Force F₄=5.0 N

Radius r = 5.0 cm

Radius R =12 cm

If we take counterclockwise torque as positive

We need to calculate the torque

Using formula of torque

[tex]\tau=RF_{1}-RF_{2}-rF_{3}+rF_{4}[/tex]

Put the value into the formula

[tex]\tau=12.0\times10^{-2}\times6.0-12\times10^{-2}\times4.0-5.0\times10^{-2}\times2.0+0\times5.0[/tex]

[tex]\tau=0.14\ N-m[/tex]

We need to calculate the moment of inertia of the cylinder

Using formula of moment of inertia

[tex]I=\dfrac{1}{2}MR^2[/tex]

[tex]I=\dfrac{1}{2}\times2.0\times(12.0\times10^{-2})^2[/tex]

[tex]I=0.0144\ kg-m^2[/tex]

We need to calculate the angular acceleration of the cylinder

Using formula of torque

[tex]\tau=I\times\alpha[/tex]

Put the value into the formula

[tex]0.14=0.0144\times\alpha[/tex]

[tex]\alpha=\dfrac{0.14}{0.0144}[/tex]

[tex]\alpha=9.72\ rad/s^2[/tex]

Positive value of acceleration shows the direction of angular acceleration is counterclockwise.

Hence, The angular acceleration of the cylinder is 9.72 rad/s^2.

batolisis

The angular acceleration is : 9.72 rad/s²

The magnitude of the angular acceleration is : 9.72

The Direction of the angular acceleration is : Counter clockwise

Given data :

Mass of cylinder = 2.0 kg , F₁ = 6.0 N ,  F₂=4.0 N ,  F₃=2.0 N ,  F₄=5.0 N

r = 0.05 m  ,  R = 0.12 m  

First step : determine the torque value

τ = RF₁ - RF₂ - rF₃ + rF₄  ----- ( 1 )

Input the given values into equation ( 1 )

τ = 0.14 N-m

Next step ; Calculate the moment of inertia of the given cylinder

I ( moment of inertia ) = [tex]\frac{1}{2} MR^{2}[/tex]   ---- ( 2 )

where ; M = 2.0 kg ,  R = 0.12 m

Input values into equation ( 2 )

I ( moment of inertia ) = 0.0144 kg-m²

Final step : Determine the angular acceleration

Angular acceleration ( Îą ) = Torque / moment of inertia

                                          = 0.14 / 0.0144

                                          = 9.72 rad/s²

Since the value of the angular acceleration is positive this shows that the direction of the angular acceleration is counterclockwise

Hence we can conclude that The angular acceleration is : 9.72 rad/s², The magnitude of the angular acceleration is : 9.72, The Direction of the angular acceleration is : Counter clockwise.

Learn more : https://brainly.com/question/21278452

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