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 ce nter 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
1 answer:
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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Answer:
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