A grindstone in the shape of a solid disk with diameter 0.520 and a mass of = 50.0 is rotating at = 860 . You press an ax against the rim with a normal force of = 250 (Figure 1) , and the grindstone comes to rest in 7.40 .
Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.
Well, it seems like this grindstone situation has come to a grinding halt! Let's see if we can find the coefficient of friction here.
First, let's find the torque exerted by the ax on the grindstone. Torque is given by the equation torque = force * radius. In this case, the force is the normal force, 250, and the radius is half the diameter, which is 0.520/2 = 0.260.
So the torque exerted by the ax is 250 * 0.260 = 65 N*m.
Next, we need to find the moment of inertia of the grindstone. The moment of inertia of a solid disk rotating about its axis is given by the equation I = (1/2) * mass * radius^2. Here, the mass is given as 50.0, so the radius is 0.260.
Substituting the values, we get I = (1/2) * 50.0 * 0.260^2 = 1.690 kg*m^2.
Since the grindstone comes to rest, we know that the torque exerted by the ax is equal to the torque due to friction, which is given by the equation torque = coefficient of friction * normal force * radius.
Substituting the values, we get 65 = coefficient of friction * 250 * 0.260.
Solving for the coefficient of friction, we get coefficient of friction = 65 / (250 * 0.260) ≈ 0.1.
So the coefficient of friction between the ax and the grindstone is approximately 0.1.
To find the coefficient of friction between the ax and the grindstone, we can use the equation for torque and the equation for rotational kinetic energy.
Step 1: Find the net torque acting on the grindstone.
The net torque is given by the equation:
net torque = moment of inertia * angular acceleration
The moment of inertia of a solid disk about its central axis is given by:
I = (1/2) * m * r^2
where m is the mass of the grindstone and r is the radius of the grindstone (half of the diameter).
Given:
Mass of the grindstone, m = 50.0 kg
Diameter of the grindstone, d = 0.520 m
Radius of the grindstone, r = d/2 = 0.520/2 = 0.260 m
Plugging the values in:
I = (1/2) * 50.0 * (0.260)^2
I = 3.38 kg·m^2
Step 2: Find the angular acceleration of the grindstone.
We can use the equation of rotational motion:
Rotational kinetic energy = (1/2) * I * (angular velocity)^2
The initial rotational kinetic energy of the grindstone is given by:
K_initial = (1/2) * I * (angular velocity_initial)^2
The final rotational kinetic energy of the grindstone is zero, as it comes to rest.
Therefore, the change in kinetic energy is:
ΔK = K_initial - 0
ΔK = K_initial
The work done on the grindstone by the friction force is given by:
Work = ΔK
The work done by the friction force is equal to the net torque times the total distance traveled by the rim.
The distance traveled by the rim of the grindstone is the circumference:
Distance = 2πr
The net torque is given by:
net torque = friction force * radius
Therefore, we have the equation:
friction force * radius * 2πr = ΔK
Simplifying the equation:
friction force = ΔK / (2πr^2)
Given:
Change in kinetic energy, ΔK = (1/2) * I * (angular velocity_initial)^2
Angular velocity_initial, ω_initial = 860 rad/s
Radius of the grindstone, r = 0.260 m
Plugging the values in:
friction force = [(1/2) * I * (angular velocity_initial)^2] / (2πr^2)
Step 3: Find the friction force.
Given:
Normal force, F_normal = 250 N
The friction force is equal to the coefficient of friction times the normal force:
friction force = coefficient of friction * normal force
Therefore, we can rearrange the equation to find the coefficient of friction:
coefficient of friction = friction force / normal force
Plugging in the values for the friction force and the normal force, we get:
coefficient of friction = [(1/2) * I * (angular velocity_initial)^2] / (2πr^2 * normal force)
Given:
Normal force, F_normal = 250 N
Plugging the values in:
coefficient of friction = [(1/2) * 3.38 * (860)^2] / (2π * (0.260)^2 * 250)
Step 4: Calculate the coefficient of friction.
Simplifying the equation:
coefficient of friction = 0.6
Therefore, the coefficient of friction between the ax and the grindstone is 0.6.
To find the coefficient of friction between the ax and the grindstone, we can use the equation for rotational motion:
Torque (τ) = Moment of inertia (I) * Angular acceleration (α)
The torque exerted on the grindstone due to the normal force and friction can be calculated as:
τ = r * F * sin(θ)
where r is the radius of the grindstone, F is the normal force, and θ is the angle between the radius and the line of action of the normal force (which in this case is 90 degrees).
Given that the grindstone comes to rest, the angular acceleration (α) can be calculated using the equation:
α = 0 - ω / t
where ω is the initial angular velocity and t is the time taken to come to rest.
Let's calculate the torque first:
r = 0.520 / 2 = 0.260 meters (since the diameter is given)
F = 250 Newtons
θ = 90 degrees = π/2 radians
τ = 0.260 * 250 * sin(π/2) = 65 N.m
Next, we can calculate the initial angular velocity (ω) using the formula:
ω = 2πf
where f is the frequency of rotation in Hertz (Hz).
Given that the frequency is 860 revolutions per minute (rpm), we need to convert it to Hz:
frequency (f) = 860 / 60 = 14.333 Hz
ω = 2π * 14.333 = 28.565 radians/second
Now we can calculate the angular acceleration:
α = (0 - 28.565) / 7.40 = -3.861 rad/s^2 (since the grindstone comes to rest, the final angular velocity is 0)
Finally, we can substitute the values into the torque equation to find the coefficient of friction (μ):
τ = I * α
I = (1/2) * m * r^2 (moment of inertia of a solid disk)
50.0 = (1/2) * 50.0 * (0.260)^2 = 0.845 kg.m^2
65 = 0.845 * (-3.861)
μ = 65 / (0.845 * 3.861)
μ ≈ 22.73
Therefore, the coefficient of friction between the ax and the grindstone is approximately 22.73.
angular momentum initial = I omega
I = (1/2) m r^2 = (1/2)(50).26^2
860 what? rpm?
860 rev/min * 1/60 min/sec * 2 pi radians/rev = 90 rad/sec
so Iw= 152
torque * time = change of angular momentum
250 N * mu * 7.4 sec = 152
I am just guessing at your units for everything