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?
The paper runs out when the child has run 40 meters. The rest is irrelevant.
40 = (1/2) at^2
80 = .35 t^2
t = sqrt (80/.35)
no
To solve this problem, we need to apply the principles of rotational motion. The key equations that we will use are:
1. The equation relating torque (τ) to moment of inertia (I) and angular acceleration (α):
τ = I α
2. The equation relating linear acceleration (a_tan) of a point on the roll to its angular acceleration (α) and its radial distance from the axis of rotation (R):
a_tan = α R
3. The equation relating angular displacement (θ) to linear displacement (s) of a point on the roll, its radial distance from the axis of rotation (R), and the total length of the paper (L):
θ = s / R = L / (2π R)
Now, let's apply these equations to solve the problem:
1. Find the angular acceleration (α):
By equation 2: α = a_tan / R = 0.35 m/s^2 / 0.075 m = 4.67 rad/s^2
2. Find the torque (τ):
By equation 1: τ = I α = (9.0 x 10^-4 kg m^2) * (4.67 rad/s^2) = 4.20 x 10^-3 N m
3. Find the total length of paper unspooled (L):
By equation 3: L = θ * (2π R) = θ * (2π * 0.075 m)
4. Find the time at which the spool runs out of paper (t):
The child maintains the same constant acceleration, so the time required to unspool the paper can be found using the kinematic equation:
L = 0.5 a_tan t^2
Rearranging the equation, we get:
t^2 = (2 L) / a_tan
Substituting the calculated value for L, we have:
t^2 = (2 * θ * (2π * 0.075 m)) / a_tan
t^2 = (2 * (L / (2π * 0.075 m)) * (2π * 0.075 m)) / a_tan
t^2 = L / a_tan
Taking the square root of both sides, we get:
t = √(L / a_tan)
Substituting the given value for L, we have:
t = √(40. m / 0.35 m/s^2) = √(114.29 s^2) = 10.70 s
Therefore, the spool will run out of paper after approximately 10.70 seconds.