Class Management | Help HW #6: Rotation and Torque Begin...

Class Management | Help HW #6: Rotation and Torque Begin Date: 4/7/2024 12:01:00 AM -- Due Date: 4/22/2024 11:59:00 PM End Date: 5/17/2024 11:59:00 PM (9%) Problem 13: Three children are riding on the edge of a merry-go-round that is a disk of mass 98 kg, radius 1.8 m, and is spinning at 19 rpm. The children have masses of 20.8 kg, 28 kg, and 35 kg. Randomized Variables M = 98 kg m? = 20.8 kg m? = 28 kg m? = 35 kg r= 1.8 m f= 19 rpm If the child who has a mass of 28 kg moves to the center of the merry-go-round, what is the new magnitude of angular velocity in rpm? f=

Transcript

00:01 Okay, so we have ourselves here, a uniform seesaw, and we have the pivot, which is to the left of the center of mass, in order to achieve equilibrium between a 21 kilogram child and 105 kilogram adult.

00:14 The adult has a mass five times the child.

00:16 The seesaw has a mass of 19 kilograms, and then what are these numbers here? well, the child and adult are offset from the ends of the seesaw by a distance.

00:31 Of 29 centimetres, this is 0 .29 meters.

00:35 This would mean that the distance to the centre from the, well, we know it's 12 metres in total, therefore, would be six from the end to the centre, six metres from the end to the centre.

00:45 However, the adult is in this case 0 .29 centimetres in, that means to the centre, this point here, will be 5 .71 metres.

00:56 5 .71 plus 0 .29 is 6.

00:59 But in order to get to the pivot, we need to go to the centre, but then backtrack, so take away this distance d, which we want to find.

01:06 Then for the child, the same idea, or a similar idea, it's 5 .71 metres to the centre, but then we need to add a distance d to get to the pivot point.

01:16 Okay, now, let's work out the clockwise and anti -clockwise talks.

01:20 I'm not going to worry about including or making the masses into forces or into weights, so employing 9 .81, the acceleration due to gravity, because it's going to be the same on both sides, it will just cancel out.

01:35 So it can just stick with the masses rather than converting it to weights and neutens.

01:38 So 105, which would take the place of our force, multiplied by the distance of the pivot, that's 5 .71 minus d, is equal to 19 times d, okay, because the centre of mass of the seesaw is distance d away from the pivot, plus the torque or the moment due to the child, that's 21 multiplied by 5 .71 plus d.

02:04 Okay, so this gets us to 599 .55 minus 105d is equal to 19d plus 21 multiplied by, sorry, i want to multiply that through, plus 119 .91 plus 21d.

02:24 Now we can move all the numbers over to one side, move all the ds over to the other side...

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