A 3.8 kg cat is standing on a rotating platform of radius of 1.5 m from the center of the platform. Then the operator turns on the ride and brings it up to a rotating rate of one revolution every 4.0 s a) Identify the net force on the cat b) Calculate the net force acting on the cat c) The cat begins to walk radially outward. He realizes that he begins to slide when he reaches a radius of 2.7 m. Assuming the same rotation rate, calculate the of static friction between the cat and the rotating platform.

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CarliReifsteck CarliReifsteck · Answer 1 · 2019-09-06T11:39:20+02:00

Answer:

(b). The net force is 14.06 N.

(c). The static friction is 0.68.

Explanation:

Given that,

Mass of cat = 3.8 kg

Radius = 1.5 m

Time = 4.0 sec

(a). We need to Identify the net force on the cat

We draw the figure for identify the net force on the cat

The net force on cat is

[tex]F=m\omega^2 r[/tex]

(b). We need to calculate the net force acting on the cat

Using formula of centripetal force

[tex]F=m \omega^2 r[/tex]

Put the value into the formula

[tex]F=3.8\times(\dfrac{2\pi}{t})^2\times r[/tex]

[tex]F=3.8\times(\dfrac{2\pi}{4})^2\times 1.5[/tex]

[tex]F=14.06\ N[/tex]

(c). We need to calculate the static friction between the cat and the rotating platform.

Using relation of frictional formula and net force

[tex]F_{\mu}=F_{net}[/tex]

[tex]\mu mg =m\omega^2 r[/tex]

[tex]\mu=\dfrac{\omega^2 r}{g}[/tex]

Put the value into the formula

[tex]\mu_{s}=\dfrac{(\dfrac{\pi}{2})^2\times2.7}{9.8}[/tex]

[tex]\mu_{s} =0.68[/tex]

Hence, (b). The net force is 14.06 N.

(c). The static friction is 0.68.