A mass M can slide with negligible friction inside a tube bent in a circle of radius R = 1.28 m. The tube rotates about its diameter; the mass is in equilibrium at θ = 50.0°. Calculate the number of revolutions the hoop makes in 108 s.
To solve this problem, we can use the concept of centripetal force.
First, let's find the acceleration of the mass M. The centripetal acceleration can be calculated using the equation:
a = R * ω^2
Where:
a is the acceleration
R is the radius of the circle
ω (omega) is the angular velocity
In this case, the angle θ is given as 50.0°. We can find the angular displacement (Δθ) in radians:
Δθ = θ * (π/180)
Substituting the given values, we have:
Δθ = 50.0° * (π/180) = 0.8727 rad
The time period can be calculated using the formula:
T = 2π/ω
Where:
T is the time period
ω (omega) is the angular velocity
The number of revolutions (N) can be found using the formula:
N = t/T
Where:
N is the number of revolutions
t is the given time period (108 s)
T is the time period
To find the angular velocity, we can use the relation between angular velocity and angular displacement:
ω = Δθ / t
Substituting the given values, we get:
ω = 0.8727 rad / 108 s = 0.00807 rad/s
Using this value of omega, we can now calculate the time period T:
T = 2π/ω = 2π / 0.00807 rad/s = 779.17 s
Finally, we can calculate the number of revolutions N:
N = t / T = 108 s / 779.17 s = 0.1389 revolutions
Therefore, the hoop makes approximately 0.1389 revolutions in 108 seconds.