How long does it take a toilet paper roll to hit the floor, if it was released f

Question

How long does it take a toilet paper roll to hit the floor, if it was released from a height of 0.73 m, while you are holding onto the first sheet? The roll has an inner radius of 2.7 cm, an outer radius of 6.1 cm, and a mass of 274 g. Assume that its mass and outer radius don’t change significantly over the course of the fall. (a) Find an expression for the acceleration of the roll depending only on the two given radii and g, through a combination of linear and rotational dynamics, then use kinematics to calculate the time. (b) Find the final speed of the roll through energy conservation, and then use kinematics to compute the time it takes the roll to hit the ground. (c) How do the results of (a) and (b) compare?

Explanation / Answer

Here,

h = 0.73 m

ri = 0.027 m

rf = 0.061 m

m = 0.274 Kg

a) as the moment of inertia is given as

I = 0.5 * m * (ri^2 + rf^2)

Using second law of motion

a = net force/effective mass

a = m * g/(m + 0.5 * m * (ri^2 + rf^2)/(rf)^2)

a = g/(1 + 0.5 *(ri^2 + rf^2)/(rf)^2)

for time taken to fall

t = sqrt(2h/a)

t = sqrt(2 * 0.73/(9.8/(1 + 0.5 *(0.027^2 + 0.061^2)/0.061^2)))

t = 0.49 s

b)

for the final speed , v

m * g * h = 0.5 * m * v^2 + 0.5 * 0.5 * m * (ri^2 + rf^2) * (v/rf)^2

putting values

9.8 * 0.73 = 0.5 * v^2 + 0.25 * (0.027^2 + 0.061^2)/0.061^2 * v^2

solving for v

v = 3 m/s

Now ,for time taken , a = v^2/(2 * d) = 3^2/(2 * 0.73)

a = 6.16 m/s^2

time taken = 3/6.16 = 0.49 s

c)

hence , the time taken for both a and b is same .

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