Solving the Angle of a Rolling Penny

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The discussion revolves around solving for the angle \Theta that a penny makes with the horizontal while rolling in a circular path. The provided formula for the angle is tan\Theta=(3v^2)/(2gR). Participants emphasize the importance of including acceleration in the free body diagram (FBD) to apply Newton's second law effectively. The centripetal acceleration, directed toward the center of the circle, is noted to have a magnitude of v^2/r. The conversation highlights the necessity of integrating all forces and accelerations to arrive at the correct solution.
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Homework Statement


If you start a penny rolling on a table with care, you can make it roll in a circle. The coin leans inward with its axis tilted. The radius of the coin is r, the radius of the circular path it follows is R, and velocity is v. Assuming that the coin does not slip, find the angle \Theta that the axis makes with the horizontal

The answer is given as
tan\Theta=(3v2)/(2gR)

I always have problems figuring out how to start the problems, but the answer usually becomes clear after I get past the first step. The only equation that I can think of that would help is I=.5mb2, but since mass isn't mentioned in the problem it seems useless.

So far what I have done is draw a free body diagram of the penny with the force of the penny's weight acting straight down, the normal force acting straight up, and frictional force directing it toward the center of the circular path it follows.

Where would be a good place to start?
 
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davesface said:
So far what I have done is draw a free body diagram of the penny with the force of the penny's weight acting straight down, the normal force acting straight up, and frictional force directing it toward the center of the circular path it follows.

Where would be a good place to start?

Hi davesface! :smile:

You've started in the right place, but you've left out the acceleration.

(you don't like acceleration, do you? :wink:)
 
I have nothing against acceleration, but that free body diagram was the hint that we got about the problem, and it did not include acceleration. I would assume that the acceleration would be toward the center of the circle, though.
 
davesface said:
I have nothing against acceleration, but that free body diagram was the hint that we got about the problem, and it did not include acceleration. I would assume that the acceleration would be toward the center of the circle, though.

Hi davesface! :smile:

An FBD can include acceleration.

(This is because an FBD works because of Newton's second law … force = mass times acceleration … so acceleration-times-mass can go in the FBD just like force :wink:)

Yes, you're right … the acceleration is toward the center of the circle ("centripetal") … its magnitude is v2/r.

Multiply it by the mass, bung it in the FBD and solve! :smile:
 
Good stuff.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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