Which one is harder to stop, a 2 kg particle moving in a circular path of radius 5 m , with angular velocity of 10 rad/s or a 2 kg particle moving in a circular path of radius 2 m with a angular velocity of 25 rad/s. I think that these two particles have the same linear momentum but their angular momentum differs, and since these are not rigid bodies and they are just particles so I think that there's no problem to deal with them using a linear analog but I crashed with this example. So what do you think?
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asked Jan 25, 2016 at 18:24
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2 Answers
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We only need worry about angular momentum when calculating other angular behaviors. In your case, you want to find the linear speed (velocity but without worrying about direction) of each ball. The speed of a point on a circle is the angular speed in radians/second multiplied by the radius of the circle, $2\pi r$ (to get to circumferences traveled per second, essentially). Once you've done that, then the linear momentum is just the speed times the mass.
BUT: if you want to stop something, you need to eliminate its kinetic energy. So if there turns out to be a difference in the masses, then be sure to calculate $\frac{mv^2}{2}$ for each case, not just $mv$.
answered Jan 25, 2016 at 20:01
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$\begingroup$ So you mean that they require the same effort to stop. $\endgroup$ Commented Jan 25, 2016 at 20:10
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$\begingroup$ @صهيبأبوريدة yes, in this case. $\endgroup$ Commented Jan 25, 2016 at 20:50
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As L=m*(r^2)*w
m= mass, r= radius, w is angular velocity
case 1 L= 2*(5^2)*10 = 500
case 2 L= 2*(2^2)*25 = 200
now off course one with larger momentum will have larger inertia when stopping hence case 1.
so no real need to go in the linear scenario.
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$\begingroup$ Wrong: all we care about is linear momentum, since angular momentum is dependent on choice of origin. Angular velocity times radius equals linear speed. $\endgroup$ Commented Jan 25, 2016 at 19:56