October 30th, 2014, 01:44 PM  #1 
Member Joined: Jul 2014 From: Seattle Posts: 96 Thanks: 2  Martial Arts Dummy  Force required question...
My friend does kickboxing and he has this dummy: http://www.amazon.com/Century%C3%82%...ial+arts+dummy You can fill the bottom with (I think it is approx) 100kg of sand in the base to make it more sturdy. I was wondering (roughly) what kind of force would be required to knock it over. Thanks for any help you can offer. 3uler. 
October 31st, 2014, 04:44 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
We can approximate the punch bag as being made up of two cylinders, a lower one with a radius of 0.6m and a height of 0.5m, which contains sand, and an upper one with a radius of 0.4m and a height of 1.5m, which contains plastic. We can also define $\displaystyle y$ as a vertical spatial coordinate where $\displaystyle y = 0$ is the floor. Density of sand = 1600 kg/m3 Density of plastic (polypropylene) = 400 kg/m3 Centre of mass of lower cylinder = $\displaystyle y_{com,1} = 0.25$ m Centre of mass of upper cylinder = $\displaystyle y_{com,2} = 1.25$ m Volume of lower cylinder = $\displaystyle \pi r^2 h = \frac{9\pi}{50}$ Volume of upper cylinder = $\displaystyle \pi r^2 h = \frac{6\pi}{25}$ Mass of lower cylinder = $\displaystyle 1600 \times \frac{9\pi}{50} = 288\pi$ Mass of upper cylinder = $\displaystyle 400 \times \frac{6\pi}{25} = 96\pi$ Position of centre of mass = $\displaystyle \frac{288\pi \times 0.25 + 96\pi \times 1.25}{288\pi + 96\pi} = 0.5$m Typically the centre of mass is within the lower part of the object, in this case the lower cylinder filled with sand. Our estimate is probably a little high, but not too out of the ordinary. We want to calculate the force required to knock it over. The force is presumably approximate to an instantaneous force, whereby we have a rapid acceleration for a short amount of time. Hence we really want to find whether a tilt of sufficient initial angular velocity can tip it over. Once the punch is made and the punch bag starts to tilt, the only force acting to correct the punch bag to its original position is gravity, which tries to tilt it back. The punch bag will fall over of its own volition if the tilt is sufficient enough so that the centre of mass is positioned above the vertex of the base and then pushed just a little more. The angle with which this will occur is $\displaystyle \arctan \left(\frac{0.3}{0.5}\right) = 31$ degrees. We will need the moment of inertia. Since I'm just doing a quick calculation, I'll approximate it as $\displaystyle I = M\left(\frac{R^2}{4} + \frac{L^2}{3}\right)$ With R = 0.5m and L = 2m. The formula above is applicable to tipping a single cylinder from its top. $\displaystyle I = 384\pi\left(\frac{1}{8} + \frac{4}{3}\right) = 560\pi$ Using rotational equivalents of Newton's laws: angular velocity = $\displaystyle \omega(t) = \omega_i  \alpha t$ where $\displaystyle \omega_i$ = initial angular velocity and $\displaystyle \alpha$ = angular acceleration Torque = $\displaystyle F \times r = I\alpha$ In the above formula, r is the perpendicular distance between the force and the corner of the base, which starts at 0.3 and reduces to zero before tipping over. F is the weight of the whole thing. The function for r with tipping angle is $\displaystyle r = 0.583\cos(\beta + 59^{\circ})$ where 0.583 is the distance between the corner and the centre of mass and $\displaystyle \beta$ is the current tipping angle (as a function of time). Consequently, $\displaystyle \omega(t) = \omega_i  \alpha t$ $\displaystyle \omega(t) = \omega_i  \frac{Fr}{I} t$ $\displaystyle \omega(t) = \omega_i  \frac{0.583F\cos(\beta+59^{\circ})}{I} t$ We now have a differential equation that we can solve to get the $\displaystyle \omega_i$ required for there to be a nonzero angular velocity at the tipping point . I'll solve it in a bit! 
October 31st, 2014, 09:18 AM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
That's awesome! I started working on this and just gave up after a certain point  too much fiddly detail to work through. I'm glad you were able to go further!

October 31st, 2014, 07:10 PM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,340 Thanks: 983 Math Focus: Wibbly wobbly timeywimey stuff.  
November 2nd, 2014, 04:20 PM  #5 
Member Joined: Jul 2014 From: Seattle Posts: 96 Thanks: 2 
Great job Benit, can you solve the differential equation? If not, can anyone else?

November 3rd, 2014, 02:28 AM  #6 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Update: In fact, the equation in my first post is rubbish because I used a formula relevant only for constant acceleration... doh! Acceleration is $\displaystyle \alpha = \frac{0.583F}{I}\cos(\beta + 59^{\circ})$ So our differential equation is $\displaystyle \frac{d^2\beta}{dt^2} = \frac{0.583F}{I}\cos(\beta + 59^{\circ})$ Last edited by Benit13; November 3rd, 2014 at 02:59 AM. Reason: Urgh. 
November 3rd, 2014, 03:32 AM  #7 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
This is very similar to that of a pendulum. For the case of a pendulum, an approximation of $\displaystyle \frac{d^2\beta}{dt^2} + a\cos (\beta + 59^{\circ}) = 0$ which is the same as $\displaystyle \frac{d^2\beta}{dt^2} + a\sin (31^{\circ}  \beta) = 0$ is $\displaystyle \frac{d^2\beta}{dt^2} + a(31^{\circ}  \beta) = 0$. Last edited by Benit13; November 3rd, 2014 at 03:35 AM. 
November 3rd, 2014, 08:04 PM  #8  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,340 Thanks: 983 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \frac{d^2\beta}{dt^2} + a \pi \left ( \frac{31^{\circ}  \beta}{180^{\circ}} \right ) = 0$ The small angle approximation to sine, $\displaystyle \sin(\theta) \approx \theta$ is true only for $\displaystyle \theta$ in radians. Dan Last edited by skipjack; November 6th, 2014 at 07:24 AM.  
November 4th, 2014, 04:19 AM  #9 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
I got a result, but it took quite a lot of calculations, so I won't specify everything on here. Solving the second order differential equation gave complex roots, (I just used 31 degrees = 0.54 radians) yielding a solution of the form $\displaystyle \beta = c_1e^{at}\sin(bt) + c_2e^{at}\cos(bt)$, where $\displaystyle a$ and $\displaystyle b$ are constants that relate to the roots of the complementary equation, $\displaystyle \beta = a \pm bi$. Using boundary conditions $\displaystyle t=0, \beta = 0, \frac{d\beta}{dt} = \omega_i$ and putting in values of $\displaystyle a$ and $\displaystyle b$ yields the specific solution $\displaystyle \beta = \frac{\omega_i}{0.417}e^{0.2t}\sin(0.417t)$. It is clear that $\displaystyle \beta$ and $\displaystyle \frac{d\beta}{dt}$ is larger for a given time $\displaystyle t$ if $\displaystyle \omega_i$ is larger. So the harder you hit it, the faster it rotates. Solving the specific solution for $\displaystyle t$ when $\displaystyle \frac{d\beta}{dt}=0$ gives $\displaystyle t = 4.84$ seconds. This is rather long, but this is the longest time it can take for the punch bag to reach the point where the centre of mass is above the vertex. Increasing $\displaystyle \omega_i$ reduces the time taken. Now we know the time taken to reach the top, we can solve for $\displaystyle \omega_i$, which is$\displaystyle \omega_i = 0.095$ radians per second. Now, a typical punch according to a random paper I looked up on google has a contact time of, perhaps, 0.1 seconds. To achieve the angular velocity of 0.095 radians per second, you need to hit the punch bag with such as force as to accelerate the punch bag by $\displaystyle a = \frac{\Delta v}{\Delta t} = \frac{r\omega_i}{\Delta }t \approx \frac{1.0 \times 0.095}{0.1} = 0.95 m/s^2 $ where I chose the punch to be at a height such that the distance between the punch contact point and the centre of mass is 1m. The mass of the whole thing is $\displaystyle 384\pi$, so the force required is $\displaystyle F = ma = 384 \pi \times 0.95 = 1146.05$ Most punches are of this order of force, so it is definitely possible to knock over the punch bag if you don't fill the bottom with enough sand. Please note that the calculations I used were really rough. Chances are the centre of mass is lower than the one I used, which puts more stringent constraints on the force required to knock it over. It wouldn't surprise me if the force required increases to something like 10 kN, which is much larger than a typical punch. Also, the force I quote above is a minimum limit. That is, the force needs to exceed that value in order to knock it over. I might try and repeat the calculation with a lower centre of mass and a larger base size. I'll post the results here when I can. 
November 4th, 2014, 04:41 AM  #10  
Member Joined: Jul 2014 From: Seattle Posts: 96 Thanks: 2  Quote:
 

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