Torque Center of Mass Lever arm 1 Lever arm 2 Lever arm...

Torque & Center of Mass Lever arm 1 Lever arm 2 Lever arm 3 1 2 3 Fig. 1 The experimental setup for torque A. Finding the center of mass (CM): A uniform meter stick has a mass of 75.0 grams and has two masses attached to it: 50.0 grams at the 20.0cm mark and 100.0 grams at the 75.0 cm mark. Find the center of mass of the system from the 0 cm point on the meter stick . B. Finding the unknown position (x) of a mass: A 50.0 grams is attached at the 20.0cm, where (x location) should you place 75 grams mass on the , meter stick to reach an equilibrium? C. Finding the unknown mass (m): A 50.0 grams is attached at the 20.0cm, what is the unknown mass (kg) that I need to place at 60 cm on the meter stick to reach an equilibrium? D. Equilibrium/balance of three masses: Three masses are attached to a uniform meter stick. The masses to the left are 50 g at 10 cm and 100 g at 40 cm. Where should you place the 200 g to reach an equilibrium? *** Draw each case above ***

Transcript

00:01 In this question, we are going to do a number of calculations regarding center of mass and balancing torques by hanging masses off of a meter stick.

00:15 So at first we are given a meter stick, we'll call it mm, of 75 grams and its center of mass is going to be right here at 50 centimeters.

00:33 And then we are told that we are hanging 50 grams at the 20 centimeter mark.

00:42 So we've hung 20 grams here.

00:49 That's actually really small.

00:50 Let me make that a little bit better.

00:55 Or, sorry, it was 50 grams.

00:56 Sorry, sorry, sorry.

00:57 It was 50 grams that we're going to hang here at the 20 centimeter mark.

01:04 And then we're going to have 100 grams at the 75 centimeter mark, which is going to be like here.

01:15 Okay.

01:18 Oh no.

01:21 There we go.

01:22 So 100 grams.

01:24 And we want to know the center of mass of this system as measured from the zero centimeter mark.

01:38 So the x position of our center of mass will be found with the meter stick mass times its position from zero plus, we'll call this one m1, m1 times x1, its position from zero.

02:01 We'll call this other one m2, plus m2 times its position from zero, all divided by the total of the masses.

02:24 So we can leave everything in grams and centimeters here.

02:30 So the first thing we're going to do is our 75 grams times our 50 centimeters.

02:37 And then we can add our 50 grams at 20 centimeters from zero plus our 100 grams at 75 centimeters from zero.

03:00 I don't draw straight lines very well when i'm doing it freehand.

03:04 There we go.

03:05 Then we're going to divide by the total of the masses, 75 plus 50 plus 100.

03:19 And i get a position of the center of the mass, the center of mass equal to 54 .4 repeating decimal centimeters.

03:32 It makes sense that we're going to have shifted it a little bit beyond the center because we've got 100 grams way out there at 75 centimeters.

03:44 And that's more than going to make up for the 50 grams a little bit closer to zero as far as sort of like balancing each other out.

03:56 Okay.

03:57 And now for part b, we are going to find the unknown position of a mass, assuming that we are balancing our meter stick at 50.

04:25 So we're assuming we're balancing our meter stick at 50.

04:28 We are going to place 50 grams at 20 centimeters.

04:39 So we'll call that one m1 and we will call this r1 where r1 is going to be 30 centimeters from 50.

04:56 Right? if we're at the 20 centimeter mark, then we are 30 centimeters away from the balance point.

05:01 And then we're going to place 75 grams or where are we going to place 75 grams to reach equilibrium? you might be able to guess that we're going to need to put the 75 grams a little bit closer to 50 in order to balance.

05:26 So we'll call this one m2.

05:30 We're going to call this a distance of r2 and we will find the actual position on the meter stick x with 50 plus r2.

05:43 And again, this m2 was 75 grams.

05:47 So we're going to set this up as a balanced torque relationship.

05:53 So we will say the torques causing counterclockwise rotation will be balancing the torques causing clockwise rotation.

06:06 And so the torque causing clockwise rotation is going to be the weight of one times its distance and the torque producing clockwise rotation will be the weight of two times its distance.

06:24 We will want to express this as kilograms and meters.

06:34 You know, let me move this over to a little more space.

06:38 We are going to want to express this in kilograms and meters.

06:45 So 0 .050 kilograms times 9 .8 for gravity times 0 .30 meters equals our 0 .075 kilograms times 9 .8 times the distance r2.

07:05 Yeah, the 9 .8 cancels, but it's important to include it so that we remember that it's not a mass that matters.

07:13 It is a force that matters.

07:20 And we are going to get 0 .20 equals r2.

07:28 Oh, come on.

07:29 There we go.

07:31 Because i'll take my 0 .05 times my 0 .3 divided by 0 .075.

07:35 And so my x position will be 0 .5 meters because it was 50 centimeters plus my 0 .20 and x is going to be putting us at 70 centimeters, 0 .70 meters.

08:03 All right.

08:03 Similar thought process for part c, which i will put down here, where we're going to have our meter stick.

08:15 I did not mean to draw it with that divot, but we're going to have our meter stick supported at 50 centimeters...

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