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A turntable has an angular velocity of 3.5 rad/s. A dust bunny is on the disk of the turntable at a distance of 0.2m from the center of rotation. The coefficient of static friction required to keep the dust bunny from getting slung off is at leastA. 0.51B. 0.45C. 0.33D. 0.12E. 0.25

Respuesta :

Answer:

E. 0.25

Explanation:

Given that

Angular speed ω = 3.5 rad/s

Distance ,r= 0.2 m

Lets take mass of dust bunny = m

We know that

Radial force F = m ω² r

The friction force on the dust bunny Fr

Fr= μ m g

To getting slung off  dust bunny from disk

m ω² r =  μ m g

ω² r =  μ  g

3.5²  x 0.2 =  μ x 10                        ( take g =10 m/s²)

μ = 0.245

μ = 0.25

Therefore answer is E

E. 0.25

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