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.**