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.
Take g=9.8m/s^2 !!!
and 0.734 is incorrect answer
To find the period T of the rolling U.S quarter, we can use the concept of rotational motion and equilibrium.
Step 1: The first thing we need to determine is the net force acting on the quarter in order to keep it in circular motion.
Since the quarter is rolling without slipping, there are two forces involved: the gravitational force acting vertically downward (mg) and the normal force (N) acting perpendicular to the inclined plane.
Step 2: To find the vertical component of the gravitational force, we need to consider the angle of inclination (θ) between the inclined plane and the horizontal.
The vertical component of the gravitational force is given by mg cos(θ).
Step 3: The normal force acting perpendicular to the inclined plane can be calculated as well.
The normal force is given by N = mg cos(θ).
Step 4: Now we need to consider the net force acting horizontally. This force is responsible for providing the necessary centripetal force to keep the quarter moving in a circle of radius R.
The net force acting horizontally is given by: N sin(θ) = (mv^2) / R, where v is the linear velocity of the quarter.
Step 5: To find the linear velocity v, we can use the relationship between linear and angular velocity. The linear velocity is equal to the angular velocity (ω) multiplied by the radius (R).
v = ωR
Step 6: The angular velocity ω can be calculated by dividing the angle (θ) by the time period (T) using the equation ω = θ / T.
Step 7: Now that we have all the necessary information, we can solve for the period T.
First, substitute v = ωR into the net force equation (from step 4):
N sin(θ) = (m(Rω)^2) / R.
Cancelling out R, we get:
N sin(θ) = mRω^2.
Substituting the values of N (mg cos(θ)) and ω (θ / T), we get:
mg cos(θ) sin(θ) = m(R(θ / T))^2.
Cancelling out "m" from both sides:
g cos(θ) sin(θ) = R^2 (θ / T)^2.
Rearranging the equation to solve for T:
T^2 = (R^2 θ^2) / (g cos(θ) sin(θ)).
Taking the square root of both sides, we have:
T = sqrt((R^2 θ^2) / (g cos(θ) sin(θ))).
Substituting the given values R = 4 cm and θ = 45°, and converting them to the appropriate units (1 cm = 0.01 m and 1° = π/180 radians), we obtain:
T = sqrt((0.04^2 * (π/4)^2) / (9.8 * cos(π/4) * sin(π/4))).
Evaluating this expression, we find:
T ≈ 0.320 seconds.
Therefore, the correct answer for the period T of the rolling U.S quarter is approximately 0.320 seconds.