A small cylinder rests on a circular turntable that is rotating clockwise at a constant speed. The
cylinder is at a distance of r = 12 cm from the center of the turntable. The coefficient of static
friction between the bottom of the cylinder and the surface of the turntable is 0.45. What is the
period of the cylinder if it is travevling at the maximum speed v
max that the cylinder can have
without slipping off the turntable?
To find the period of the cylinder, we need to consider the forces acting on the cylinder and determine the condition for it to stay on the turntable without slipping off.
First, let's identify the forces acting on the cylinder:
1. The gravitational force (mg), acting vertically downward.
2. The normal force (N), which is equal in magnitude but opposite in direction to the gravitational force, acting vertically upward.
3. The static friction force (fs), acting horizontally inward towards the center of the turntable (opposite to the radial direction).
To determine the maximum speed (v_max) without slipping, we need to ensure that the static friction force is equal to the maximum value it can reach, given by the equation:
fs = μs * N
where μs is the coefficient of static friction between the cylinder and the turntable.
Since the cylinder is at rest on the turntable, the magnitude of the gravitational force (mg) is balanced by the magnitude of the normal force (N). Therefore, we have:
mg = N
Now, let's consider the forces in the radial direction (horizontal):
The centripetal force required to keep the cylinder moving in a circle is given by:
Fc = m * (v^2 / r)
where m is the mass of the cylinder, v is its velocity, and r is the distance from the center of the turntable.
Since the static friction force (fs) provides the necessary centripetal force (Fc), we have:
fs = Fc
Substituting the expressions for fs and Fc into the equation and rearranging, we get:
μs * N = m * (v^2 / r)
Now, substituting N = mg, the equation becomes:
μs * mg = m * (v^2 / r)
Canceling out mass (m) from both sides, we get:
g * μs = v^2 / r
Simplifying further, we find:
v^2 = g * μs * r
Finally, taking the square root of both sides, we get:
v = √(g * μs * r)
To calculate the period (T) of the cylinder, we have the formula:
T = 2πr / v
Substituting the expression for v, we get:
T = 2πr / √(g * μs * r)
Plugging in the given values:
r = 12 cm = 0.12 m
μs = 0.45
g = 9.8 m/s^2
T = 2π * 0.12 / √(9.8 * 0.45 * 0.12)
Calculating further, we find:
T ≈ 2.54 seconds
Therefore, the period of the cylinder is approximately 2.54 seconds.
To find the period of the cylinder, we need to first determine the maximum speed at which it can rotate without slipping off the turntable. We can use the static friction coefficient and the distance of the cylinder from the center of the turntable to calculate this.
The maximum speed that the cylinder can have without slipping off the turntable occurs when the static friction force between the cylinder and the turntable is at its maximum. This is given by the formula:
F_friction = μ * N
where F_friction is the frictional force, μ is the coefficient of static friction, and N is the normal force.
The normal force acting on the cylinder can be calculated as the product of the mass of the cylinder and the acceleration due to gravity:
N = m * g
where m is the mass of the cylinder and g is the acceleration due to gravity.
Since the cylinder is at a distance of r = 12 cm from the center, we can use the centripetal force equation to relate the maximum speed v_max to the normal force N:
N = m * g = m * (v_max)^2 / r
Rearranging this equation, we can solve for v_max:
(v_max)^2 = r * g
v_max = sqrt(r * g)
Substituting the given values:
v_max = sqrt(0.12 m * 9.8 m/s^2) = sqrt(1.176) m/s = 1.084 m/s
Now, we can calculate the period of the cylinder using the formula:
T = 2π * r / v_max
Substituting the values:
T = 2π * 0.12 m / 1.084 m/s = 0.698 s
Therefore, the period of the cylinder traveling at the maximum speed that it can have without slipping off the turntable is approximately 0.698 seconds.
mv²/R =μmg
v=sqrt(μgR) =
=sqrt(0.45•9.8•0.12)=0.73 m/s
T=2πR/v = 2π•0.12/0.73 = 1.03 s.