A potter's wheel having a radius 0.45 m and a moment of inertia of 14.9 kg · m2 is rotating freely at 49 rev/min. The potter can stop the wheel in 8.0 s by pressing a wet rag against the rim and exerting a radially inward force of 72 N. Find the effective coefficient of kinetic friction between the wheel and the wet rag.
49 rev/min * 1 min/60 s * 2 pi radians/rev
= 5.13 radians/s
angular acceleration alpha = 5.13 rad/s / 8 s
= .641 radians/s^2
M = torgue or moment = I alpha
= 14.9 * .641 = 9.56 Newton meters
F = M/radius = 9.56/.45 = 21.2 Newtons
so
21.2 = mu * 72
mu = .295
To find the effective coefficient of kinetic friction between the wheel and the wet rag, we can analyze the forces acting on the wheel.
The key concept here is that the external force exerted by the potter using the wet rag causes a torque that opposes the angular momentum of the spinning wheel.
First, let's convert the given rotational speed of the wheel from revolutions per minute (rev/min) to radians per second (rad/s).
Angular velocity (ω) = 2π × (revolutions/time in seconds)
= 2π × (49/60)
= 51.3 rad/s
Next, let's calculate the initial angular momentum of the wheel before any external force is applied.
Angular momentum (L) = moment of inertia (I) × angular velocity (ω)
= 14.9 kg · m^2 × 51.3 rad/s
= 763.37 kg · m^2/s
Now, let's determine the change in angular momentum when the potter stops the wheel.
Change in angular momentum (ΔL) = Final angular momentum (Lfinal) - Initial angular momentum (Linitial)
= 0 - 763.37 kg · m^2/s
= -763.37 kg · m^2/s
To calculate the torque exerted by the potter:
Torque (τ) = Force (F) × Lever Arm (r)
= 72 N × 0.45 m
= 32.4 N · m
The negative sign indicates that the torque acts in the opposite direction of the initial angular momentum, as the force is applied radially inward.
Now, let's find the relationship between torque and change in angular momentum:
Torque (τ) = Change in angular momentum (ΔL) / Time (t)
= -763.37 kg · m2/s / 8.0 s
= -95.42 kg · m^2/s^2
Finally, let's calculate the effective coefficient of kinetic friction using the equation for torque:
τ = Iα
Where α is the angular acceleration and is related to the torque and moment of inertia as follows:
α = τ / I
Plugging in the values:
-95.42 kg · m^2/s^2 = 14.9 kg · m^2 × α
Simplifying:
α = -6.41 rad/s^2
The negative sign indicates that the angular acceleration acts in the opposite direction of the initial angular velocity.
Now, let's calculate the net torque acting on the wheel due to the frictional force:
τnet = Iα
τnet = 14.9 kg · m^2 × (-6.41 rad/s^2)
= -95.21 kg · m^2/s^2
We know that the net torque is also equal to the force of friction (Ff) multiplied by the lever arm (r), so:
Ff × r = -95.21 kg · m^2/s^2
To find the force of friction (Ff), we divide both sides of the equation by the lever arm (r):
Ff = -95.21 kg · m^2/s^2 / 0.45 m
= -211.58 N
The negative sign indicates that the force of friction is acting radially outward.
The effective coefficient of kinetic friction (μ) can be calculated using the equation:
Ff = μ × N
Where N is the normal force, which is equal to the weight of the wheel.
To find the weight of the wheel:
Weight (W) = Mass (M) × Acceleration due to gravity (g)
Assuming the mass of the wheel is negligible compared to the mass of the potter's wheel, we can calculate the weight:
Weight (W) = 0.45 m × 9.8 m/s^2
= 4.41 N
Now, let's plug in the values to calculate the effective coefficient of kinetic friction:
-211.58 N = μ × 4.41 N
Simplifying:
μ = -211.58 N / 4.41 N
= -48
Since the coefficient of kinetic friction cannot be negative, there may be an error in the calculation or given values. Please double-check the values provided and ensure the appropriate units are used in the calculation.
To find the effective coefficient of kinetic friction between the wheel and the wet rag, we need to use the equation that relates torque and rotational motion.
The equation is:
Torque = Moment of inertia x Angular acceleration
Since the potter can stop the wheel, the angular acceleration is negative (opposite direction of rotation). We need to find the angular acceleration in order to determine the torque exerted by the radially inward force.
To find the angular acceleration, we can use the equation:
Angular acceleration = (Final angular velocity - Initial angular velocity) / Time
Given:
Radius (r) = 0.45 m
Moment of inertia (I) = 14.9 kg·m^2
Initial angular velocity (ωi) = 49 rev/min
Final angular velocity (ωf) = 0 (since the wheel stops)
Time (t) = 8.0 s
First, let's convert the initial angular velocity from rev/min to rad/s:
ωi = (49 rev/min) x (2π rad/rev) x (1 min/60 s) = 49 x 2π / 60 rad/s
Now we can calculate the angular acceleration:
Angular acceleration = (0 - ωi) / t
Next, we need to find the torque exerted by the radially inward force (τ):
τ = I x Angular acceleration
Finally, we can find the effective coefficient of kinetic friction (μk) using the equation:
τ = Frictional force x Radius
Rearranging the equation to solve for μk:
μk = τ / (Frictional force x Radius)
Given:
Force (F) = 72 N
Now plug in the values and calculate:
ωi = (49 x 2π) / 60 rad/s = 8.14 rad/s
Angular acceleration = (0 - ωi) / t = (0 - 8.14 rad/s) / 8.0 s = -1.01875 rad/s^2
τ = I x Angular acceleration = 14.9 kg·m^2 x (-1.01875 rad/s^2) = -15.183125 N·m
μk = τ / (Frictional force x Radius) = -15.183125 N·m / (72 N x 0.45 m) ≈ -0.470
Therefore, the effective coefficient of kinetic friction between the wheel and the wet rag is approximately -0.470.