The radius of a roll of paper is 7.6 cm and its moment of inertia is I = 2.9 × 10^3 kg·m2. A force of 3.2 N is exerted on the end of the roll for 1.3 s, but the paper does not tear so it begins to unroll. A constant friction torque of 0.11 m·N is exerted on the roll which gradually brings it to a stop. Assume that the paper's thickness is negligible.
a.) Calculate the length of paper that unrolls during the time that the force is applied (1.3 s)
b.) the length of paper that unrolls from the time the force ends to the time when the roll has stopped moving
To answer both parts of the question, we need to consider the rotational motion of the roll of paper.
a.) To calculate the length of paper that unrolls during the time the force is applied:
1. Calculate the initial angular velocity (ω0) using the formula: τ = Iα, where τ is the torque and α is the angular acceleration. Since the force is applied at the end of the roll, the torque can be calculated by τ = rF, where r is the radius and F is the force.
τ = (7.6 cm)(3.2 N) = 24.32 N·cm
24.32 N·cm = (2.9 × 10^3 kg·m2) α
Solve for α: α = (24.32 N·cm) / (2.9 × 10^3 kg·m^2) = 0.0084 rad/s^2
2. Calculate the final angular velocity (ω) after 1.3 seconds using the equation: ω = ω0 + αt
ω = 0 + (0.0084 rad/s^2)(1.3 s) = 0.0109 rad/s
3. Calculate the angle rotated (θ) using the equation: θ = ω0t + (1/2) αt^2
θ = 0 + (1/2)(0.0084 rad/s^2)(1.3 s)^2 = 0.0076 rad
4. Calculate the length of paper unrolled (L) using the formula: L = rθ
L = (7.6 cm)(0.0076 rad) = 0.058 cm
Therefore, the length of paper that unrolls during the time that the force is applied is 0.058 cm.
b.) To calculate the length of paper that unrolls after the force ends until the roll has stopped:
1. Calculate the time (t') it takes for the roll to come to a stop using the equation: τ = Iα', where τ is the friction torque and α' is the deceleration.
τ = (7.6 cm)(0.11 mN) = 0.836 mN·cm
0.836 mN·cm = (2.9 × 10^3 kg·m^2) α'
Solve for α': α' = (0.836 mN·cm) / (2.9 × 10^3 kg·m^2) = 2.88 × 10^-4 rad/s^2
2. Calculate the time (t") it takes for the roll to stop using the equation: ω = α't"
ω = 0.0109 rad/s (since this is the final angular velocity from part a)
0.0109 rad/s = (2.88 × 10^-4 rad/s^2) t"
Solve for t": t" = 0.0109 rad/s / (2.88 × 10^-4 rad/s^2) = 37.99 s
3. Calculate the angle (θ') covered during the time the roll comes to a stop using the equation: θ' = ωt' - (1/2) α't'^2
θ' = (0.0109 rad/s)(37.99 s) - (1/2)(2.88 × 10^-4 rad/s^2)(37.99 s)^2 = 0.194 rad
4. Calculate the length of paper unrolled (L') using the formula: L' = rθ'
L' = (7.6 cm)(0.194 rad) = 1.475 cm
Therefore, the length of paper that unrolls from the time the force ends until the roll has stopped moving is 1.475 cm.
To solve both parts of the problem, we will use the equation relating torque, moment of inertia, and angular acceleration:
τ = I * α
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
a) To find the length of the paper that unrolls during the time the force is applied, we first need to calculate the angular acceleration using the torque exerted during that time.
The torque exerted on the roll is given by the force applied multiplied by the radius:
τ = F * r = 3.2 N * 0.076 m = 0.2432 N·m
Using the equation τ = I * α, we can solve for α:
0.2432 N·m = (2.9 × 10^3 kg·m^2) * α
Solving for α:
α = 0.2432 N·m / (2.9 × 10^3 kg·m^2) ≈ 8.4 × 10^-5 rad/s^2
Now, we can use the angular acceleration to find the angular displacement during the 1.3 seconds the force is applied:
θ = α * t = 8.4 × 10^-5 rad/s^2 * 1.3 s ≈ 1.09 × 10^-4 radians
The length of paper that unrolls can be calculated using the formula for the arc length:
s = r * θ = 0.076 m * 1.09 × 10^-4 radians ≈ 8.28 × 10^-6 m ≈ 8.28 μm
Therefore, during the time the force is applied, approximately 8.28 μm of paper unrolls.
b) To calculate the length of paper that unrolls from the time the force ends to the time when the roll has stopped moving, we need to find the time it takes for the roll to come to a stop.
The friction torque acting on the roll is given as 0.11 m·N. Since the friction torque is constant, it can be used to find the angular acceleration when the force is no longer applied.
Using τ = I * α, we can solve for α:
0.11 N·m = (2.9 × 10^3 kg·m^2) * α
Solving for α:
α = 0.11 N·m / (2.9 × 10^3 kg·m^2) ≈ 3.79 × 10^-5 rad/s^2
To find the time it takes for the roll to stop, we can use the equation of motion for rotational motion:
ω = ω0 + α * t
Where ω is the initial angular velocity, ω0 is the final angular velocity when the roll stops, α is the angular acceleration, and t is the time taken.
Since the initial angular velocity is not given, we assume that it is zero (since the roll is at rest when the force is applied).
So, ω0 = 0.
Now we can solve for t:
0 = α * t
t = 0 / α = 0
The time taken for the roll to stop is 0 seconds.
Therefore, no paper unrolls from the time the force ends to the time when the roll has stopped moving.
67
τ t = ΔIω
Solve for ω
θ = ωt
and s = rθ
Same technique for part two with new τ