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A 5 × 10–6-kg dot of paint on the side of a rotating cylinder flies off when the angularspeed of the cylinder reaches 5 × 103 rad/s. The spin axis of the cylinder is vertical andits radius is 0.04 m. The force of adhesion between the paint and the surface isapproximatelyA) 1 NB) 1 mNC) 5 mND) 5 kNE) 5 N

User Ipinak
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1 Answer

5 votes

Answer:

E) 5 N

Step-by-step explanation:

The dot of paint on the side of the rotating cylinder experiences an pseudo force called centrifugal force in radially outward direction.

The centrifugal force on a rotating body of mass m, angular velocity w and radius r is given by the expression.

Centrifugal force,


F=mrw^(2)

When this centrifugal force F just exceeds adhesive force between the paint and the surface the dot flies off.

from the question we get the value angular velocity at which the dot flies of as
5X10^(3)

we have,


m= 5 X10^(-6)

r = 0.04 m

therefore the centrifugal force at this point,

F = mr
w^(2)

=
(5X10^(-6) ) X (5X10^(3) )^(2) X0.04

= 125 X 0.04

= 5 N

so, Adhesive force or Adhesion = 5 N

User Rojalin Sahoo
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