英锐光电2016年营收2258万元 业绩亏损222万元

Question

Lately I observed that the bars that is used to do bench press in gym don't immediately fall on the side loaded with weight. Which does not sound correct intuitively, the bar should fall down because there is more weight on one side. So I figured that this must be related to balancing the torque exerted by the weight and the torque exerted by the bar. Here is my approach:

Assume the length of the bar is $L_0$, the mass of the bar is $m$, the mass of the weight is $m_w$, the distance between the weight and the pivot is $L_w$, the distance between the end of the bar to the pivot is $L$, here is a diagram I draw to clarify what I mean.

I first started by computing the rotational inertia of this system, the rotational inertia contributed by the weight is $I_w = m_wL_w^2$, by Parallel Axis Theorem, the rotational inertia of the bar is $I_{bar} = I_{CM} + md^2 = \frac{1}{12} m L_0^2 + m (\frac{1}{2}L_0-L)^2$, the rotational inertia of the system is thus $$I_{sys} = m_wL_w^2 + \frac{1}{12} m L_0^2 + m (\frac{1}{2}L_0-L)^2$$

Now it is the torque, the torque applied by the weight is $\tau_w = m_w gL_w$, I encountered a little bit of trouble when calculating the torque of the bar, but I think this is how to do it:

the torque applied by the two yellow part should cancel out because they are symmetrical to the pivot, thus the net torque should be applied in the middle of the blue part, assume the mass is uniformly distributed throughout the bar and denote linear mass density as $\lambda$, the torque applied by the bar is $$\tau_{bar} = \lambda (L_0-2L)g\left(\frac{L_0 - 2L}{2} + L\right)$$

Thus the total net torque is $$\tau_{net} = m_wgL_w - \frac{\lambda g L}{2}(L_0-2L)$$ The angular acceleration of the system is $$\alpha = \frac{\tau_{net}}{I_{sys}}$$ Here I substitute the numerical value ($L_0 = 2.3$ m, $m = 20$ kg, $m_w = 20$ kg, $L_w = 0.05$ m, $L = 0.2$m ) to compute $\alpha$ and got $|\alpha| = 6.687 $ rad/s$^2$, which is way to big for it to stay stationary.

This result does not match the reality. Where did I made a mistake? Are there any other simpler ways to tackle this problem?

Answer 1

I gave this problem on a midterm exam one time, after having the same thought as you while at the gym. I think you are overcomplicating things considerably. In particular, I don't see what one gains by dividing the bar into yellow and blue pieces, and I don't think the moment of inertia of the bar is relevant. (Either it rotates or not. Who cares how fast?)

Here is the problem statement:

A $2.2\,\mathrm{m}$ long barbell with mass $m_B = 20\,\mathrm{kg}$ rests on two supports. The supports are each $50\,\mathrm{cm}$ from the ends of the bar. A plate with mass $m$ is placed on the right side of the barbell $40\,\mathrm{cm}$ from the end. You can assume the center of gravity of the barbell is halfway between its ends. What is the maximum mass $m$ that can be placed at this position before the barbell begins to tip?

Here is the solution:

The net torque taken about the right support is \begin{align*} \tau_{\mathrm{net}} = \tau_L + \tau_B + \tau_m \end{align*} where $\tau_L$ is the torque due to the left support, $\tau_B$ is the torque due to the weight of the bar, and $\tau_m$ is the torque due to the plate. All the forces involved are vertical, so the moment arms are just the horizontal distances from the pivot point, and we get \begin{align} \tau_{\mathrm{net}} = -(1.2\,\mathrm{m}) N_L + (0.60\,\mathrm{m}) m_B g - (0.10\,\mathrm{m}) m g \end{align} In order for the bar to remain is static equilibrium, that is, for it not to tip, we need $\tau_{\mathrm{net}} = 0$, so \begin{align} N_L = \frac{1}{12} m g - \frac{1}{2} m_B g \end{align} When the bar begins to tip, the normal force from the left support will go to zero, so if the bar does not tip, we have $N_L > 0$, which means \begin{align} \frac{1}{12} m g &- \frac{1}{2} m_B g > 0 \\ \rightarrow m &< 6 m_B \end{align} So as long as the mass of the plate is less than six times the mass of the bar, the bar won't tip over.

I tried to estimate relatively realistic numbers (without actually taking my tape measure to the gym) but of course you're welcome to adjust them as necessary. Obviously, adding more plates rather than swapping out with a single heavier plate changes the effective moment arm, but the calculation works the same.