A governor (centrifugal pendulum) consists of two 200–g spheres attached by light, rigid 10.0–cm rods to a vertical rotating axle. The rods are hinged so that the spheres swing out from the axle as they rotate with it. However, when the angle θ is 45.0°, the spheres encounter the wall of the cylinder in which the governor is rotating (see the figure above). (a) What is the minimum rate of rotation, in revolutions per minute, required for the spheres to touch the wall? (b) If the coefficient of kinetic friction between the spheres and the wall is 0.35, how much retarding force results as the spheres rub against the wall when the mechanism is rotating at 300 rev⁄min?
To find the minimum rate of rotation required for the spheres to touch the wall, we can start by considering the motion of one of the spheres. At the point where the sphere touches the wall, the gravitational force acting on it provides the necessary centripetal force to keep it moving in a circular path.
(a) To find the minimum rate of rotation, we can equate the gravitational force acting on the sphere with the centripetal force:
mg = mω²r
Here, m is the mass of the sphere, g is the acceleration due to gravity, ω is the angular velocity in radians per second, and r is the distance between the axle and the wall (10.0 cm or 0.10 m in this case).
The mass of each sphere is given as 200 g, so we need to convert it to kilograms:
m = 200 g = 0.200 kg
The acceleration due to gravity is approximately 9.8 m/s².
Now, we can solve for the angular velocity ω by rearranging the equation:
ω² = (mg) / (mr)
ω² = (0.200 kg * 9.8 m/s²) / (0.10 m)
ω² = 19.6 rad/s²
ω = sqrt(19.6 rad/s²)
ω ≈ 4.43 rad/s
To convert the angular velocity to revolutions per minute, we can use the conversion factor:
1 rev/min = (2π rad) / (60 s)
ω (in rev/min) = (4.43 rad/s) * (60 s / (2π rad))
ω ≈ 134 rev/min
So, the minimum rate of rotation required for the spheres to touch the wall is approximately 134 revolutions per minute.
(b) To find the retarding force when the mechanism is rotating at 300 rev/min, we can use the coefficient of kinetic friction to determine the frictional force acting on the spheres. The frictional force will oppose the motion of the spheres.
The frictional force can be calculated using the equation:
frictional force = coefficient of kinetic friction * normal force
The normal force is the gravitational force acting on the sphere, which is given by:
normal force = m * g
Substituting the given values:
normal force = (0.200 kg) * (9.8 m/s²)
normal force ≈ 1.96 N
Now, we can calculate the frictional force:
frictional force = (0.35) * (1.96 N)
frictional force ≈ 0.686 N
So, when the mechanism is rotating at 300 rev/min, the retarding force resulting from the spheres rubbing against the wall is approximately 0.686 N.
To answer part (a) of the question, we need to find the minimum rate of rotation, in revolutions per minute (rpm), required for the spheres to touch the wall when the angle θ is 45.0°.
To solve this, we can use the concept of centripetal force. At the moment the spheres touch the wall, the centripetal force acting on each sphere is equal to the weight of the sphere. This can be expressed as:
F_c = m * g
Where F_c is the centripetal force, m is the mass of each sphere (200 g = 0.2 kg), and g is the acceleration due to gravity (9.8 m/s^2).
The centripetal force can also be expressed as:
F_c = m * r * ω^2
Where r is the radius of rotation and ω is the angular velocity. The angular velocity can be converted to revolutions per minute (rpm) as follows:
ω = 2π * n / 60
Where n is the number of revolutions per minute.
Since the spheres are attached by light, rigid rods of 10.0 cm, the radius of rotation is half the rod length, or 5.0 cm = 0.05 m.
Setting the two expressions for F_c equal to each other, we can solve for the minimum rate of rotation:
m * r * ω^2 = m * g
Simplifying the equation:
r * ω^2 = g
Plugging in the values:
0.05 * (2π * n / 60)^2 = 9.8
Solving for n (rate of rotation):
n = sqrt((9.8 * 60^2) / (2π * 0.05))
Calculating this value will give us the minimum rate of rotation, in revolutions per minute, required for the spheres to touch the wall.
To answer part (b) of the question, we need to find the amount of retarding force that results as the spheres rub against the wall when the mechanism is rotating at 300 rev/min (rpm).
To solve this, we can use the concept of friction. The retarding force can be calculated using the equation:
F_friction = μ * F_N
Where F_friction is the friction force, μ is the coefficient of friction (0.35), and F_N is the normal force between the spheres and the wall.
The normal force, F_N, can be calculated as:
F_N = m * g * cos(θ)
Where m is the mass of each sphere (0.2 kg), g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle at which the spheres encounter the wall (45.0°).
Plugging in the values:
F_N = 0.2 * 9.8 * cos(45.0°)
Once we have the normal force, we can calculate the friction force:
F_friction = 0.35 * (0.2 * 9.8 * cos(45.0°))
Calculating this value will give us the amount of retarding force that results as the spheres rub against the wall when the mechanism is rotating at 300 rev/min.