Trang chủ » blog » SOLVED:A coin placed 30.0 \mathrm{~cm} from the center of a rotating, horizontal turntable slips when its speed is 50.0 \mathrm{~cm} / \mathrm{s}. (a) What provides the force in the radial direction w

SOLVED:A coin placed 30.0 \mathrm{~cm} from the center of a rotating, horizontal turntable slips when its speed is 50.0 \mathrm{~cm} / \mathrm{s}. (a) What provides the force in the radial direction w

Video Transcript

Hello and welcome to this television solution of enumerate. This question is based on a circular motion. here we have a mint that is placed at the concentrate of a rotating horizontal turntable. so from the center at the distance of the coin from the center of it as art which is equal to 30 centimeters now this is equal 2.3 m. And the coin actually slips when the rapid ‘s 50 centimeters was second. now when the turntable is rotating at 30 50 let us draw something like this. This is a turntable and this is a center and you have a coin out hera. now. This turntable is rotating with sudden speed V. And that when we equal to 30 to win the equal 50 centimeter was irregular, that is equal 2.5 molarity per second. then this coin begins to sleep. Hope what now ? What provides the military unit in the radial focus when the mint is stationary proportional to the town ? Okay, so when the coin was stationary there were two forces. I think the the first part forces. The centripetal is a centripetal force. nowadays this centripetal wedge is generated ascribable to the rotation of this mint about sudden center and it is directed towards the center. So we can say the centripetal storm is directed towards the plaza of this act to them and look sol the mint is stationary. So there must be some outbound pull that is besides opposing the centripetal military unit that actually keeps this coin stationary. therefore this reverse forces nothing but the frictional power. So what the wonder it is asked only. What is that radial effect ? So it ‘s the centripetal force ? So option A. We have centripetal forces, answer no B in B. It is asked. What is the coefficient of electrostatic friction between the coin and the timetable ? So we should balance these two forces. The centripetal push, let ‘s say F. C. I do not as a centrifugal coerce which is he called to the frictional force. 1/4 immediately, centripetal forces given way aims every squarely by our and which is equal to meal meter, 17 maintain the slant of this uh mint. so hera we have the mask gets canceled. We need to calculate mu. So you equal to we squared by jesus. now here we have the square S 0.5 feather by Gs 9.8 times the radius of rotation with this point. That keeps us zero 0.8 sol 0.8 is the coefficient of friction between this mint and the turntable. I hope this is clear to you and have a very dear rest of the sidereal day. Thank you.

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