You have a grindstone (a disk) that is 98.0 kg, has a 0.335-m radius, and is turning at 100 rpm, and you press a steel axe against it with a radial force of 23.6 N. Assuming the kinetic coefficient of friction between steel and stone is 0.192, calculate the angular acceleration of the grindstone.

Respuesta :

Answer:

[tex]a=0.276[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=98.kg[/tex]

Radius [tex]r=0.335[/tex]

Angular velocity [tex]\omega=100rpm[/tex]

Radial force of [tex]F_r=23.6 N.[/tex]

Kinetic coefficient of friction  [tex]\mu=0.192[/tex]

Generally the equation for Kinetic Force is mathematically given by

 [tex]F_k=\mu.F_r[/tex]

 [tex]F_k=0.192*23.6[/tex]

 [tex]F_k=4.5312[/tex]

Generally the equation for Torque on Center is mathematically given by

 [tex]Ia=f_k*r[/tex]

Where

 [tex]I=\frac{Mr^2}{2}[/tex]

Therefore

 [tex]a=\frac{2f_k}{Mr}[/tex]

 [tex]a=\frac{2*4.5312}{98*0.335}[/tex]

 [tex]a=0.276[/tex]

Therefore Angular acceleration of the grindstone is

[tex]a=0.276[/tex]

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