Explanation: two toilet paper rolls are dropped such that they land on the floor at the same time. One roll of toilet paper is unravelling as it falls from the height of the desk (while one person holds the toilet paper to the desk as the roll falls and continues to unravel), and the other roll is falling from an unknown height, that is of course higher than the unravelling roll, so that both rolls land on the floor at the same time. The attached picture contains all the information on the objects, visualizing the scenario.
Roll 1 is falling from the table and unravelling as it falls (it has a mass of 0.10129kg)
Roll 2 is falling from the unknown height to meet roll 1 on the ground at the same time (this has a mass of 0.09884kg)
The height of the desk from which roll 1 is falling from is 0.915m.
The measured diameter of the rolls across the entire thing = 10.5cm (0.105m), which gives a radius of 5.25cm (0.0525m).
The measured diameter of the rolls across the hollow cardboard center = 5.5cm (0.055m), which gives a radius of 2.25cm (0.0225m).
The measurements were equal on each roll.
These rolls are treated as hollow thick-walled cylindrical objects, such that their moment of inertia `I ` is given by the equation
`I = 1/2 M (r^2 + R^2) ` where M is the mass, r is the inner radius and R is the outer radius
Calculate the height from which roll 2 must fall in order that it reach the floor at the same time as roll 1.
NB The change in total radius of the unravelling roll, roll 1, from desk to floor is assumed to be negligible compared to its original length R, so that the moment of inertia for the roll, and indeed the roll's mass, is approximately constant throughout its fall.
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Assume H=height for dropped roll (roll 2) (to be determined), h=height for unrolled roll (roll 1) = 0.915m , r = inner radius = 0.0225m , R = outer radius = 0.0525m, M = mass of unrolled roll = 0.10129kg
First, since the dropped roll (roll 2) is dropped from a stationary position (initial velocity = 0) we have that, by standard Newton mechanics (kinematics)
`H = 1/2 g t_(drop)^2 ` (1)
where `t_(drop) ` is the time the rolls both take to drop (given as equal). The acceleration acting on the falling roll is just that due to gravity (`g = 9.8m^(-2) ` )
For the unrolling toilet paper roll (roll 1), similarly
`h = 1/2 a t_(drop)^2 ` (2)
but where the linear (downwards, towards the ground) acceleration acting on it, `a `, is determined by the rolling action (rotational motion) of the falling roll.
From these two equations we have the relation that
`H/h = g/a ` ,giving that
`H = h(g/a) ` (3)
To calculate the linear acceleration `a ` of the unrolling roll (roll 1), we first note the equation for the moment of inertia for a thick-walled cylinder:
`I = 1/2 M (r^2 + R^2) `
However, this needs to be adapted (using the parallel-axis theorem) to account for the unrolled roll rotating about its outer radius `R ` . This gives the moment of inertia of the unrolling roll as
`I = 1/2 M (r^2 + R^2) + MR^2 = 1/2 M (r^2 + 3R^2) `
Next, using a free body diagram to identify the net torque on the roll as `tau_("net") = MgR ` , and using Newton’s 2nd Law for Rotational Motion to find the angular acceleration `alpha ` , namely that
`tau_("net") = I/alpha `
we can put these two together to obtain
`alpha = tau_("net")/I = (MgR)/(1/2 M (r^2 + 3R^2)) = (2gR)/(r^2 + 3R^2) `
Now, since the linear acceleration, `a `, of the unrolling roll (roll 1) can be found from its angular acceleration `alpha ` multiplied by the radius of rotation (`R ` here, the outer radius), that is
`a = alpha R = (2gR^2)/(r^2 + 3R^2) ` (4)
then, finally, using the formula (3) above for the height to be calculated` ``H `and substituting
the value for the linear acceleration `a ` given by (4) , we obtain
`H = (hg(r^2 + 3R^2))/(2gR^2) = (h(r^2 + 3R^2))/(2R^2) `
which here is given by
`H = (0.915(0.0225^2 + 3(0.0525^2)))/(2(0.0525^2)) = 1.456531` m
The height `H ` of the dropped roll (roll 2), that drops under no torque force, is then approximately 1.46m to 3 significant figures.
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