A 14.0-kg solid sphere of radius 1.60 m is being rotated by applying a constant tangential force of 10.0 N at a perpendicular distance of 1.60 m from the rotation axis through the center of the sphere. If the sphere is initially at rest, how many revolutions must the sphere go through while this force is applied before it reaches an angular speed of 35.0 rad/s?

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Respuesta :

aristeus aristeus · Answer 1 · 2019-11-13T07:56:14+01:00

Answer:

Angular displacement will be 548.8 rad

Explanation:

We have given mass of the solid sphere m = 14 kg

radius R = 1.60 M

We know that moment of inertia is given by

[tex]I=\frac{2}{5}mR^2=\frac{2}{5}\times 14\times 1.6^2=14.336kgm^2[/tex]

Force is given by F = 10 N

And perpendicular distance is given as d = 1.6 m

We know that torque is equal to [tex]\tau=Fd=I\alpha[/tex]

[tex]10\times 1.6=14.336\times \alpha[/tex]

[tex]\alpha =1.11rad/sec^2[/tex]

Now final angular velocity [tex]\omega =35rad/sec[/tex]

From third equation of motion we know that

[tex]\omega ^2=\omega _0^2+2\alpha \Theta[/tex]

[tex]35 ^2=0^2+2\times 1.116\times \Theta[/tex]

[tex]\Theta =548.8rad[/tex]