Texts a merry go round is initially at rest has a radius of...

Name: texts a merry go round is initially at rest has a radius of 21 m and a mass of 125 kg a child of mass 34 kg sits on the very edge of the merry go round while another child exerts a constant 89375
Uploaded: 2022-04-28T10:56:04-08:00
Duration: 6 min 29 s
Channel: Sri K
Description: texts a merry go round is initially at rest has a radius of 21 m and a mass of 125 kg a child of mass 34 kg sits on the very edge of the merry go round while another child exerts a constant 89375

Texts: A merry-go-round is initially at rest, has a radius of 2.1 m, and a mass of 125 kg. A child of mass 34 kg sits on the very edge of the merry-go-round while another child exerts a constant force of 10 N to the outside of the merry-go-round in the clockwise direction tangential to the merry-go-round. (a) What is the moment of inertia of the merry-go-round? (Hint: treat it as a disk/cylinder.) (b) What is the moment of inertia of the child? (Hint: he is much smaller than the radius of the merry-go-round and thus can be treated as a point mass.) (c) What is the net torque on the merry-go-round? (d) What is the angular acceleration of the merry-go-round/child system? (e) How long does it take the child to travel a physical distance of 100 m? What is his angular displacement at this time? (f) What is the angular velocity at this time?

Transcript

00:01 Here given that a child of mass 25 kilograms is moving on the edge of a merry -go -round and radius of this is 2 meters and mass of mary -go -round wheel is 150 kilogram whose angular velocity omega initial is equal to 1 revolution per second is equal to 2 pi radian per second.

00:31 Now we know that for finding out a part and we have to calculate a second.

00:36 Initial moment of inertia so for this it will be half m r squared plus m r square so here it will be half that means 0 .5 m is m of the wheel that is 150 and here it is 2 square plus it will become 25 and r is 2 square so it will become here 400 kilo meter per meter square that is i initial and here we have to tell angular momentum l i that is equal to i initial and omega initial so i omega that means 400 multiply by 2 5 so which is equal to 2513 kilogram meter square per second so the both answer is here for the key part one is this this part as l i and here this part now for further moving to the b part so we now we have to tell final moment of inertia i f will be equal to half m r square which is equal to 0 .5 and m is as 150 multiply by 2 square so which is equal to 300 kilogram meter square that is i final and for same as angular momentum that is l final angular momentum so it is again equals to l i because both are conserved so here lf will be final is equals to 2 513 kilogram meter square per second and here we see these both the answers s lf is equal to 2 ,513 and if 300 kilogram meter per second square.

03:38 Now we see here c part.

03:43 We have to find it out use l final is equal to i final and omega which is omega final is equal to l final divided by i final which is equal to 2513 and divided by so omega final is equal to 8 .38 radian per second or 1 .33 revolution per second now we will see the d part which is as omega f is equals to 8 .38 gradient per second or 1 .33 revolution per second...

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