A coin is placed 15cm from the axis of a rotating turntable of variable speed. w

Question

A coin is placed 15cm from the axis of a rotating turntable of variable speed. when the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 31 rpm is reached, at which point the coin lides off. what is the coefficient of static friction between the coin and the turntable?

Please help answer the followin questions. I am a visual learner, so I would appreciate it if there is hand written work showing the steps.

4. -110 points Giancoli7 5.P 015 My Notes Ask Your Teacher 0 A coin is placed 15.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 31 rpm is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable? 5. 10 points Giancoli? 5.P.018. My Notes Ask Your Teacher Tarzan plans to cross a gorge by swinging in an arc from a hanging vine. If his arms are capable of exerting a force of 1700 N on the rope, what is the maximum speed he can tolerate at the lowest point of his swing? His mass is 78 kg and the vine is 4.8 m long. m/s

Explanation / Answer

(1) The Centrifugal force will make the coin slide off the turntable.

And for the coin to slide off, the centrifugal force must be able over come the force of friction trying to hold off the coin from slipping.

Fc = centrifugal force = m*V^2 / r ....................(1)

where,

m = mass of the coin
V = velocity of the the turn table
r = turning radius = 15 cm.

w = rotational speed of turntable = (31 rev/min)*(2 pi rad/rev)*(1 min/60 sec)

w = 3.246 rad/sec.

Therefore,

V = 3.246 * 0.15 = 0.486 m/sec.

Substituting this in Equation 1,

Fc = m(0.486)^2 / (0.15)

Fc = 1.577* m N

Frictional force = (mu)*(m)*(g)

where

mu = coefficient of friction
m = mass of coin
g = acceleration due to gravity = 9.8 m/sec^2

Substituting values,

1.577*(m) = (mu)*(m)*(9.8)

Since "m" appears on both sides of the equation, it will cancel out and solving for "mu",

mu = 1.577 / 9.8

mu = 0.161

(2)

at lowest point > being at the botton of circlular arc of verticle cirecle > he will face following set of forces

m*v^2 /r = centripetal = upwards

T = tension > always acting on tied point = upwards

weight = mg = down

so

mv^2 / r = T - mg = 1700 - 78*9.8

v^2 = 4.8[1700 - 78*9.8] / 78

v = 7.58 m/s = at lowest point of his swing

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