A disk having moment of inertia $$100kg\cdot m^2$$ is free to rotate without friction, starting from rest, about a fixed axis through its center. A tangential force whose magnitude can range from \(F = 0\) to \(F = 50.0\) N can be applied at any distance ranging from \(R = 0\) to \(R = 3.00\) m from the axis of rotation. Find a pair of values of \(F\) and \(R\) that cause the disk to complete 2.00 rev in 10.0 s. Choose the pair with the greatest force.
F R = I alpha
alpha = change in angular velocity omega/ time
if angular acceleration alpha is constant ( constant torque)
omega = alpha t
angle = (1/2) alpha t^2
so
2 pi * 2 = (1/2) alpha t^2
alpha t^2 = 8 pi = 25.1 radians
t = 10 seconds
so
alpha = 25.1/100 = 0.251 radians/s^2
therefore F R = 100 * 0.251 = 25.1 Newton meters
max F = 50
so R = 25.1 / 50 or about 1/2 meter
To find the pair of values of \(F\) and \(R\) that cause the disk to complete 2.00 rev in 10.0 s, we have to consider the angular acceleration of the disk.
The formula for the angular acceleration can be derived from the torque equation:
\[\tau = I\alpha\]
where:
\(\tau\) is the torque,
\(I\) is the moment of inertia, and
\(\alpha\) is the angular acceleration.
For a disk rotating freely without friction, the torque equation simplifies to:
\[\tau = I\alpha\]
The torque can be calculated as the product of the applied force (\(F\)) and the distance from the axis of rotation (\(R\)):
\[\tau = FR\]
Substituting this into the torque equation, we get:
\[FR = I\alpha\]
Since the disk starts from rest and completes 2 revolutions in 10.0 seconds, we can calculate the angular acceleration (\(\alpha\)) using the equation:
\[\theta = \frac{1}{2}\alpha t^2\]
where:
\(\theta\) is the angular displacement (in radians),
\(t\) is the time, and
\(\alpha\) is the angular acceleration.
The angular displacement can be calculated by converting 2 revolutions into radians:
\[\theta = 2\pi \times 2\]
Now we can find the angular acceleration:
\[\theta = \frac{1}{2}\alpha t^2\]
\[2\pi \times 2 = \frac{1}{2}\alpha (10.0)^2\]
Simplifying the equation gives:
\[4\pi = 50\alpha\]
\[\alpha = \frac{4\pi}{50}\]
Now we can find the pair of values for \(F\) and \(R\) by rearranging the torque equation:
\[FR = I\alpha\]
\[F = \frac{I\alpha}{R}\]
Plugging in the given values of \(I\), \(\alpha\), and \(R\):
\[F = \frac{100 \, \text{kg} \cdot m^2 \cdot \frac{4\pi}{50}}{3.00 \, m}\]
Simplifying the expression further:
\[F = \frac{200\pi}{3} \approx 209.4 \, \text{N}\]
Therefore, the pair of values that cause the disk to complete 2.00 rev in 10.0 s with the greatest force is \(F = 209.4\) N and \(R = 3.00\) m.
To solve this problem, we need to use the rotational analogs of Newton's laws and kinematic equations.
The moment of inertia (I) represents an object's resistance to rotational motion. For a disk rotating about its center, the moment of inertia is given by \(I = \frac{1}{2} m r^2\), where \(m\) is the mass of the disk and \(r\) is the radius.
In this case, the given moment of inertia of the disk is 100 kg·m². Since the disk is rotating about its center, we can write \(I = \frac{1}{2} m r^2\) as \(100 = \frac{1}{2} m (3)^2\) (using the given value of \(R = 3.00\) m). Solving for \(m\), we find \(m = \frac{100}{\frac{1}{2} \cdot 3^2} = 6.67\) kg.
Now, we need to find the torque (τ) applied to the disk, which is equal to the product of the force (F) applied tangentially to the disk and the distance (r) from the axis of rotation. We can write \(τ = F \cdot R\).
To determine the force and distance that cause the disk to complete 2.00 revolutions (2 complete rotations) in 10.0 s, we need to use the rotational analog of the kinematic equation:
\(θ = ω_0 t + \frac{1}{2} α t^2\),
where \(θ\) is the angle traversed in radians, \(ω_0\) is the initial angular velocity, \(t\) is the time, and \(α\) is the angular acceleration.
In this case, the disk starts from rest, so \(ω_0 = 0\). We want to find the force and distance that cause the disk to rotate 2.00 revolutions (2 × 2π radians). The time is given as 10.0 s. Since the disk starts from rest, the angular acceleration can be determined using the equation \(ω = ω_0 + α t\), where \(ω\) is the final angular velocity. Since we want the disk to complete 2 rotations in 10.0 s, we have \(ω = \frac{2 \cdot 2π}{10} = \frac{4π}{10}\) rad/s. Substituting the known values into the equation, we have \(ω = 0 + α \cdot 10.0\), which gives \(α = \frac{4π}{10 \cdot 10.0}\) rad/s².
Now, using the equation \(θ = ω_0 t + \frac{1}{2} α t^2\) with the given values, we can calculate the angle traversed (\(θ\)). We want the disk to complete 2 revolutions, so \(θ = 2 \cdot 2π\) radians. Substituting the known values, we have \(2 \cdot 2π = 0 + \frac{1}{2} \cdot \frac{4π}{10 \cdot 10.0} \cdot (10.0)^2\), which simplifies to \(θ = 2 \cdot 2π = \frac{4 \cdot π}{10} \cdot 100\). Solving for \(θ\), we find \(θ = \frac{4π \cdot 100}{10} = 40π\) radians.
To find the applied force and distance, we use the equation \(θ = \frac{τ}{I}\), where \(τ\) is the torque and \(I\) is the moment of inertia. We rearrange the equation to solve for the torque:
\(τ = θ \cdot I\).
Substituting the known values, we have \(τ = 40π \cdot 100\).
Finally, we substitute the torque equation \(τ = F \cdot R\) with the calculated torque to solve for the force (F):
\(40π \cdot 100 = F \cdot R\).
Since we want to select the pair with the greatest force, we need to maximize the force. To do so, we need to find the maximum value of \(R\) that satisfies the equation. In this case, the maximum value of \(R\) is given as 3.00 m. Thus, we can write \(40π \cdot 100 = F \cdot 3.00\) and solve for \(F\):
\(F = \frac{40π \cdot 100}{3.00}\).
Calculating this expression, we find:
\(F = \frac{4000π}{3}\) N.
Therefore, the pair of values that cause the disk to complete 2.00 revolutions in 10.0 s with the greatest force is \(\left(\frac{4000π}{3}\, \text{N}, 3.00\, \text{m}\right)\).