Respuesta :
Answer:
Explanation:
Radius of circular path of coin R = 12.5 x 10⁻²,
coefficient of static friction μs = .33
In order that the coin rotates in circular path , it requires centripetal force which is provided by friction. As speed of rotation increases , force of friction also increases to provide it required centripetal force. When the speed of rotation becomes too high so that frictional force can not compensate the increase in centripetal force then coin will start slipping or it starts moving with respect to turntable.
b ) At this point of time
centripetal force = limiting force of friction
mω² R = μs mg , m is mass of the coin , ω is angular velocity ,
ω² R = μs g
ω² x 12.5 x 10⁻², = .33 x 9.8
ω² = .33 x 9.8 / 12.5 x 10⁻²,
= 25.87
ω = 5.08 rad / s
2π / T = 5.08
T = 2π / 5.08
= 1.23 s .
The coin starts to slip when the centripetal force exceeds the static frictional force after 1.23s
In order that the coin rotates in a circular path, a centripetal force is required. This centripetal force is actually a frictional force. When the speed of rotation becomes very high the static frictional force can not balance the centripetal force because the static friction force has a limiting value proportional to the weight of the body. The coin will start slipping on the turntable.
(b) At this point of time
centripetal force = force of friction
mω² R = μ(s)mg
where m is mass of the coin, ω is angular velocity
Radius of circular path of coin R = 12.5 x 10⁻²,
coefficient of static friction μ(s) = .33
ω² R = μ(s)g
ω² x 12.5 x 10⁻², = .33 x 9.8
ω² = .33 x 9.8 / 12.5 x 10⁻²,
= 25.87
ω = 5.08 rad / s
2π / T = 5.08
T = 2π / 5.08
T = 1.23s is the after which the coin starts to slip.
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