Question

In: Physics

A coin is moving in a horizontal circle in a cone. The cone has height 70...

A coin is moving in a horizontal circle in a cone. The cone has height 70 cm and radius 20 cm. The coin is at a height halfway up. The coefficient of static friction is 0.3. The mass of the coin is 5 g. How fast is the coin moving to stay at this height?If the velocity is halved what will the height of the coin be now?

Expert Solution

In triangle ABC

tan = BC/AB

tan = r/h

tan = 20/70

tan = 0.286

= tan^-1(0.286)

= 16 deg

m = mass of coin = 5 g = 0.005 kg

u_s = Coefficient of friction = 0.3

f_s = static frictional force = u_s F_n

Along the vertical direction, force equation is given as

F_n Sin + f_s Cos = mg

F_n Sin + (u_s F_n ) Cos = mg

F_n = mg /(Sin + u_s Cos)

F_n = (0.005 x 9.8) /(Sin16 + (0.3) Cos16)

F_n = 0.087 N

Along the horizontal direction , force equation is given as

F_n Cos - f_s Sin = m v²/r

F_n Cos - u_s F_n Sin = m v²/r

0.087 Cos16 - (0.3 x 0.087)Sin16 = (0.005) v²/0.20

v = 1.75 m/s

genius_generous answered 4 months ago

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