A U.S quarter is rolling on the floor without slipping in such a way that it describes a circular path of radius R=4 cm. The plane of the coin is tilted at an angle of θ=45∘ with respect to the horizontal plane (see the figure below). Find the coin's period T in seconds, that is, the time it takes for the coin to go around the circle of radius R. The radius of a U.S quarter is r=1.2 cm.
Newton’s 2nd law
vector(ma) = vector(mg ) +vector( N) +vector(Ffr)
Projections:
ma=F(fr)
0=N-mg
mv²/r =μN
N=mg
mv²/r = kN = μmg
μ=v²/gr.
tanα =F(fr)/N = μmg/mg = μ = v²/gr.
v=sqrt{gr•tanα}
T=2πR/v= 2πR/ sqrt{gr•tanα} = …
nope, 0.733 is wrong
wait, Elena, what is are R and r? Are they the same ones that I used?
R=0.04 m
t= 0.012 m
The solution is correct.May be there is mistake in calculations (or in your answer)...
net torque on the coin, answer not just force balance
To find the coin's period T, we need to use the formula for the period of a particle moving in a circular path:
T = 2πr / v,
where T is the period, r is the radius of the circle, and v is the speed of the particle.
In this case, the radius of the circle is R = 4 cm, and the speed of the particle (the rolling quarter) is related to its rotational speed ω and the radius of the coin r by the equation v = ωr.
To find ω, we need to determine the angular speed of the coin. Since the coin is rolling without slipping, the linear speed of a point on the edge of the coin must be equal to the product of the angular speed and the radius. In other words, v = ωr.
Substituting this value of v in the formula for the period, we get:
T = 2πr / (ωr).
Now we need to find the value of ω. To do this, we can use the relationship between the linear speed v and the angular speed ω:
v = ωr.
But we are given the angle θ at which the plane of the coin is tilted with respect to the horizontal plane. This angle θ is related to the angular speed ω by the equation:
θ = ωt,
where t represents time.
Since the coin takes T seconds to go around the circle, we can write:
θ = ωT.
Rearranging this equation, we can solve for ω:
ω = θ / T.
Substituting this value of ω into the expression for the period T, we get:
T = 2πr / (θ / T * r).
Simplifying this expression, we have:
T^2θ = 2π.
Finally, we can solve for the period T:
T = sqrt(2π / θ).
Substituting the given value of θ = 45° (converting to radians: θ = π/4), we get:
T = sqrt(2π / (π/4))
= sqrt(8)
≈ 2.83 seconds.
Therefore, the coin's period T is approximately 2.83 seconds.