A student holds a bike wheel and starts it spinning with an initial angular speed of 7.0 rotations per second. The wheel is subject to some friction, so it gradually slows down. In the 10-s period following the inital spin, the bike wheel undergoes 60.0 complete rotations.
Assuming the frictional torque remains constant, how much more time Δts will it take the bike wheel to come to a complete stop?
The bike wheel has a mass of 0.825 kg and a radius of 0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque τf that was acting on the spinning wheel.
To find the time it will take for the bike wheel to come to a complete stop, we can use the concept of angular deceleration.
Step 1: Calculate the angular deceleration (α):
We can use the formula for angular deceleration:
α = (final angular velocity - initial angular velocity) / time
We are given that the initial angular velocity (ωi) is 7.0 rotations per second and the final angular velocity (ωf) is 0 rotations per second. The time (t) is 10 seconds (the 10-second period following the initial spin).
α = (0 - 7.0) / 10 = -0.7 rotations per second squared
Step 2: Find the time it will take to come to a complete stop (Δts):
Since the angular deceleration (α) remains constant, we can use the equation:
Δθ = ωi * Δt + (1/2) * α * (Δt)^2
Since we want the wheel to come to a complete stop (Δθ = 360 degrees or 2π radians), we can rearrange the equation:
0 = ωi * Δt + (1/2) * α * (Δt)^2
Solving this quadratic equation will give us the time it takes for the wheel to stop.
In our case, ωi = 7.0 rotations per second and α = -0.7 rotations per second squared.
0 = 7.0 * Δt - 0.35 * (Δt)^2
Simplifying the equation further:
0.35 * (Δt)^2 - 7.0 * Δt = 0
Since we are looking for a positive time, the solution to this equation will be:
Δt = (-(-7.0) ± sqrt((-7.0)^2 - 4 * 0.35 * 0)) / (2 * 0.35)
Δt = (7.0 ± sqrt(49.0)) / 0.7
Δt = (7.0 ± 7.0) / 0.7
Δt = 14.0 / 0.7 = 20.0 seconds
Therefore, it will take an additional 20.0 seconds for the bike wheel to come to a complete stop (Δts).
Now, let's move on to finding the magnitude of the frictional torque (τf) that was acting on the spinning wheel.
Step 1: Calculate the moment of inertia (I) of the bike wheel:
The moment of inertia (I) of a solid cylindrical object rotating about its central axis is given by the formula:
I = (1/2) * m * r^2
Where m is the mass of the wheel and r is the radius.
For our case, m = 0.825 kg and r = 0.315 m.
I = (1/2) * 0.825 * (0.315)^2
I = 0.0415 kg * m^2
Step 2: Calculate the frictional torque (τf):
The frictional torque (τf) is given by the equation:
τf = I * α
Using the values we calculated earlier, I = 0.0415 kg * m^2 and α = -0.7 rotations per second squared:
τf = 0.0415 * -0.7
τf = -0.0289 N * m
Note: The negative sign indicates that the frictional torque opposes the direction of motion.
Therefore, the magnitude of the frictional torque (τf) acting on the spinning wheel is 0.0289 N * m.
To find the answer to the first question, we need to calculate the rate at which the bike wheel is slowing down. This can be done by finding the angular acceleration using the equation:
α = (ωf - ωi) / t
where:
α is the angular acceleration
ωf is the final angular speed (which is 0 in this case since the wheel comes to a stop)
ωi is the initial angular speed (given as 7.0 rotations per second)
t is the time interval (given as 10 seconds)
Using these values, we can calculate α:
α = (0 - 7.0) / 10 = -0.7 rotations per second squared
From here, we can use the equation for angular displacement:
θ = ωi * t + (1/2) * α * t^2
where:
θ is the angular displacement (given as 60 complete rotations)
ωi is the initial angular speed (given as 7.0 rotations per second)
α is the angular acceleration (calculated as -0.7 rotations per second squared)
t is the time interval between the initial and final angular speeds (unknown)
Substituting the known values, we get:
60 = 7.0 * t + (1/2) * (-0.7) * t^2
Simplifying this equation, we have a quadratic equation in terms of t:
0.35t^2 - 7t + 60 = 0
Solving this equation, we find two possible values for t:
t = 5.2 seconds (ignoring the negative solution)
Therefore, it will take an additional 5.2 seconds for the bike wheel to come to a complete stop (Δts = 5.2 s).
To find the magnitude of the frictional torque τf, we can use the equation:
τf = I * α
where:
τf is the frictional torque
I is the moment of inertia of the wheel
α is the angular acceleration (calculated as -0.7 rotations per second squared)
The moment of inertia of a solid disk rotating about its central axis is given by:
I = (1/2) * m * r^2
where:
m is the mass of the wheel (given as 0.825 kg)
r is the radius of the wheel (given as 0.315 m)
Substituting the values, we get:
I = (1/2) * 0.825 * 0.315^2 = 0.052 kg·m^2
Finally, substituting the values into the equation τf = I * α, we get:
τf = 0.052 * (-0.7) = -0.0364 N·m
The magnitude of the frictional torque acting on the spinning wheel is 0.0364 N·m.
Correction: 70-10 = 60 Addition seconds to stop.
Vo = 7.0rev/s.
V = 60rev/10s = 6rev/s.
V = Vo + a*t.
6 - 7 + a*10, a = -0.1 rev/s^2.
V = Vo + a*t.
0 = 7 - 0.1*t, t = 70 s. = Stopping time.
70 - 6 = 64 additional seconds to stop.