# Torque Approximations Neglecting Friction

Disclaimer: I'm not asking for homework help. I don't understand how my calculations work out conceptually. If you can help me understand the connection between the concepts, I'd really appreciate it.

The following equation is used to determine the torque ($τ=mgr_{drum}$) and this is not totally correct because in principle the hanging mass still has acceleration. For the trial with maximum $α$, calculate the linear acceleration (${a = αr_{drum}}$) and then determine the exact torque (${τ{_{ex}}}$). Is the approximation for torque ($τ=mgr_{drum}$) justified?

The maximum $α$ value is 0.123 rad/$s^2$. Thus a = (0.123 rad/$s^2$)(0.50 m) = 0.0615 m/$s^2$. $τ_{ex}$= $m (g – a)$ so $τ_{ex}$ = (0.1 kg)(9.8 – 0.0615) = 0.97 N. The torque using the approximation is 0.49 N. Percent error is |0.49-0.97|/0.97 * 100% ≈ 49.5%. I don't think is is a good approximation because of the large percent error and using other alpha values, the error is high. Since the radius is constant, it doesn't affect acceleration. So, ${τ_{ex}}$ shouldn't be any different from the $τ=mgr_{drum}$ approximation given. But my large percent error contradicts this. I don't understand why they are so different.

• But $a$ seems so small compared to $g$. How could such a small $a$ throw your answer off that much? Your answers must be different for another reason. – Brian Moths Jun 26 '15 at 21:24
• Try setting the $a$ equal to 0 instead of .0615 m/s^2 and see if the two torques agree. – Brian Moths Jun 26 '15 at 21:26
• When a = 0 T_ex = 0.98, and T_approx = 0.49. Percent error is 50%. – imaginov Jun 26 '15 at 21:47
• But shouldn't $T_{ex}$ and $T_{approx}$ be the same when $a=0$. How are they different? – Brian Moths Jun 26 '15 at 21:54
• Well the radius is accounted for in T_approx (0.50) so T_ex is 2*T_approx. – imaginov Jun 26 '15 at 22:03

I am going describe the experiment that your question concerns, then I will restate your question, and then I will resolve your question and explain what you did wrong and how you could avoid make similar mistakes in the future.

# The experiment

The experiment in question deals with rotational motion. The purpose of the experiment is to measure the rotational inertia of a drum using rotational dynamics. First I will describe the set-up of the experiment, then I will describe how to find the moment of inertia of the drum.

The experiment set-up is composed of a drum with string wrapped around it and a falling weight connected to the string, as shown in the above figure. The rotational axis of the drum is fixed in space, but the drum is free to rotate about this axis. The weight is free to fall under the influence of gravity, except that the mass is attached to the inextensible string wrapped around the drum. Thus, as the mass falls, the drum must unspool the string, meaning that the drum must rotate.

Now if the mass of falling weight and the angular acceleration of the drum is measured, then it is possible to deduce the drum's moment of inertia. This is achieved through the formula $\tau = I \alpha$, where $\tau$ is the torque on the drum, $I$ is moment of inertia of the drum, and $\alpha$ is the drum's angular acceleration. To deduce the moment of inertia $I$, it is necessary to know $\alpha$ and $\tau$. As stated in the first sentence of the paragraph, $\alpha$ is measured directly, but $\tau$ is not; besides $\alpha$, only the mass $m$ of the falling weight is known.

Thus, $\tau$ must be calculated from $\alpha$ and $m$. To see how, first consider what happens if the drum prevented from rotating. Then the drum cannot spool out any string, so the mass cannot fall. In this case, the tension $T$ in the string is given by $T=mg$, and so the torque $\tau$ on the drum from the spring is $\tau = TR = mgR$, where $R$ is the radius of the drum.

Now consider what happens when the drum is allowed to rotate. In contrast to the previous case, the drum can spool out string, and so the mass does fall. If the drum has angular acceleration $\alpha$, then, since the string is inextensible, the linear acceleration $a$ of the hanging weight is given by $a=\alpha R$. Now since the hanging weight is accelerating, the tension $T$ of the rope must not be exactly balancing the gravitation force $mg$; rather, the tension must be less than the gravitational force: $T<mg$. The amount of $T$ from $mg$ is $ma$ by newton's second law to be : $T=mg - ma$. The torque can be calculated from $T$ through the formula $\tau = TR$ as before. Thus we find $\tau = (mg -ma)R = m(g-a)R$.

Now we know how to find both $\tau$ and $\alpha$, so that we may deduce the moment of inertia $I$.

We have found the formula $\tau = m(g-a)R$, but a simplifying approximation can be made: if the acceleration of the falling weight $a$ is very small compared to $g$, then $m(g-a)R \approx mgR$ and so we can approximate $\tau \approx mgR$. To distinguish the exact torque from the approximate torque, I will denote the exact torque with $\tau_e$: $\tau_e = m(g-a)R$. I will denote the approximate torque by $\tau_a$: $\tau_a = mgR$.

You were trying to calculate both $\tau_e$ and $\tau_a$ and see if the approximation $\tau_a \approx \tau_e$ did in fact hold. You found that the approximation was bad, even though you expected it to be a good approximation. (I think you had a completely wrong reason for expecting it to be a good approximation, but I won't go into that.) Your question is basically where did you go wrong.
We have to look at what you did. First you calculated $\tau_a$ as $m g R$. Plugging in $m=.1 kg$, $g=9.8 m/s^2$ and $R=.5 m$, you found $\tau_a=.49 N\cdot m$ (Note: you actually just give units of Newtons, which is wrong). Next you have to compute $\tau_e$. You started by computing $a$ as $\alpha R$. Having found $a$, your next step was to compute $\tau_e$ as $m(g-a)$. You found $\tau_e = .97 N$, which is about a factor of two off from your $\tau_a$ value.
The reason that values disagree is simply that you used the wrong formula for $\tau_e$. You used $\tau_e =m(g-a)$ when you were supposed to use $\tau_e = m(g-a)R$. The fact that $R$ has the value of $.5 m$ explains why your value was off by a factor of two, and why your answer had units of $N$ instead of $N \cdot m$.
An even better way to avoid making errors would be to work with symbols until you obtain a simplified expression for your final answer, and then and only then plug in numbers. For example, in this problem, you would write the relative error in the approximation as $\dfrac{|\tau_a - \tau_e|}{\tau_e}$. Plugging in the definitions you find the relative error is $\dfrac{|mgR - m(g-a)R|}{m(g-a)R} = \dfrac{a}{g-a}$. Now you can use $a = \alpha R$, and plug this $a$ into the formula. Since the formula $\dfrac{a}{g-a}$ is a lot simpler than $\dfrac{|mgR - m(g-a)R|}{m(g-a)R}$, it should be a lot harder to make a mistake.