A roll of toilet paper (a partially hollow cylinder with R_2 = 7.5cm, M = 300g, and I = 9.0 x 10^-4 kg m^2) is mounted on a mass-less axle along its central axis. Te roll is initially at rest. Then, at t=0, a child grabs the end of the roll and starts running, pulling the paper off the roll at a constant linear acceleration a_tan = 0.35 m/s^2.
*Assume that the roll has uniform density. Throughout this question, assume that both M and R_2 remain constant, even though paper is unspooling from the roll.*
QUESTION: The spool holds only 40. m of paper. If the child maintains the same constant acceleration, at what time will the spool run out of paper?
------- Where do I begin with this problem? -----
From the other parts of the problem I found out that the inner radius, R_1 is 0.019m (1.9 cm) and that the magnitude of the torque acting on the spool is .0042 N m (not sure if I did my work here correctly.) What equations would I have to use to solve the problem?
To solve this problem, you will need to use the equation for angular acceleration, which is given by:
α = τ/I
where τ is the torque acting on the spool and I is the moment of inertia of the spool.
You can then use the equation for angular velocity, which is given by:
ω = αt
where α is the angular acceleration and t is the time.
Finally, you can use the equation for linear velocity, which is given by:
v = ωR
where ω is the angular velocity and R is the radius of the spool.
Using these equations, you can calculate the time it will take for the spool to run out of paper. First, calculate the angular acceleration using the equation α = τ/I. Then, calculate the angular velocity using the equation ω = αt. Finally, calculate the linear velocity using the equation v = ωR. The time it will take for the spool to run out of paper is equal to the linear velocity divided by the linear acceleration (v/a_tan).
To solve this problem, you can use the kinematic equations and the equation for linear acceleration. Here are the steps to follow:
1. Calculate the total length of paper initially on the roll.
Total length = 2πR2
2. Determine the final radius Rf when the spool runs out of paper.
Rf = sqrt((M - m)/π)
3. Find the linear acceleration at which the paper is unspooling.
a_lin = R2 * a_tan
4. Using the equation of motion, find the time it takes for the spool to run out of paper.
R1 + (R2 - R1) - Rf = (1/2) * a_lin * t^2
Now let's substitute the known values into these equations and solve the problem step by step.
Given:
R2 = 7.5 cm = 0.075 m
M = 300 g = 0.3 kg
I = 9.0 x 10^-4 kg m^2
a_tan = 0.35 m/s^2
Total length = 40 m (given)
1. Total length of paper initially on the roll:
Total length = 2πR2 = 2π(0.075) = 0.47 m
2. Determine the final radius Rf when the spool runs out of paper:
Rf = sqrt((M - m)/π) = sqrt((0.3 - (0.47 / 2π))/π) = sqrt(0.1/π) ≈ 0.178 m
3. Find the linear acceleration at which the paper is unspooling:
a_lin = R2 * a_tan = 0.075 * 0.35 = 0.02625 m/s^2
4. Using the equation of motion, find the time it takes for the spool to run out of paper:
R1 + (R2 - R1) - Rf = (1/2) * a_lin * t^2
R1 - Rf = (1/2) * a_lin * t^2
(0.019 - 0.178) = (1/2) * 0.02625 * t^2
-0.159 ≈ 0.013125 * t^2
t^2 = -0.159 / 0.013125 ≈ -12.1212
t ≈ sqrt(-12.1212)
The negative result indicates that the spool does not run out of paper.
Therefore, based on the given values and the calculation, the spool does not run out of paper.
To solve this problem, you can start by using the equation relating torque (τ) and moment of inertia (I) to angular acceleration (α):
τ = Iα
Since the roll is initially at rest and the linear acceleration is known, you can relate the linear acceleration (a_tan) and angular acceleration (α) using the equation:
a_tan = R_2α
where R_2 is the outer radius of the roll.
By substituting R_2α for a_tan in the torque equation, we get:
τ = I(a_tan / R_2)
Next, you can use the fact that the torque τ is equal to the force F multiplied by the radius of the roll (R_2), and the force F is equal to the mass (M) of the unspooled paper multiplied by the linear acceleration (a_tan). Therefore, you get:
τ = F * R_2
I(a_tan / R_2) = F * R_2
From this equation, you can solve for F:
F = (I * a_tan) / R_2^2
Now, since F = M * a_tan, you can set up an equation relating M, a_tan, and the linear acceleration due to the child's pulling:
M * a_tan = (I * a_tan) / R_2^2
Solving for M, you get:
M = I / R_2^2
Given that M and R_2 are constant, this means that the mass of the remaining unspooled paper remains constant throughout the process.
Now, to find the time at which the spool runs out of paper, you know that the length L of the paper unrolled at any given time t is related to the linear acceleration (a_tan) using the equation:
L = (1/2) * a_tan * t^2
You are given that L = 40 m, and you need to find the time t when the spool runs out of paper.
Using the equation above, you can set up the following equation:
40 = (1/2) * a_tan * t^2
Rearranging and solving for t, you get:
t^2 = (80 / a_tan)
t = √(80 / a_tan)
Substituting the given value for a_tan (0.35 m/s^2), you can calculate the time at which the spool runs out of paper.
Please note that in this explanation, I assumed that the unrolling paper hangs down vertically from the roll and that no slipping occurs between the roll and the paper.