Calculating the Torque for a Hand-Driven Merry-Go-Round

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To calculate the torque required for a hand-driven merry-go-round accelerating from rest to 18 RPM in 13 seconds, the moment of inertia (I) of the system must include the mass of the disk and the two children sitting on it. The uniform disk's moment of inertia is calculated as I = 0.5 * M * R^2, resulting in 3609 kg m². The angular acceleration (alpha) is determined to be approximately 0.144997 rad/s². The torque is then calculated using the formula torque = I * alpha, yielding a value of 523 N·m, which was deemed incorrect. The discussion emphasizes the need to account for the additional mass of the children in the total moment of inertia for accurate torque calculations.
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A dad pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 18 in 13.0 . Assume the merry-go-round is a uniform disk of radius 3.0 and has a mass of 760 , and two children (each with a mass of 21 ) sit opposite each other on the edge.

Calculate the torque required to produce the acceleration, neglecting frictional torque.

What force is required at the edge?



Homework Equations



torque= I alpha
w=2pi f
w=w0+alpha *t

The Attempt at a Solution



change f to w
w= 2pi *.3rev/sec w=1.8849
find alpha
w=w0+ alpha*t 1.8849=alpha*13 alpha = .144997

find I
I= .5 MR^2= 3609
torque= I alpha = 3609*.144997= 523 this answer was incorrect I do not know why I also tried it with I for a hoop the answer was 1046 but this answer was also incorrect if you can tell me where i messed but that would be very helpful thanks
 
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Hi kaite, welcome to PF.

Please add units to all given numerical quantities and restate the problem. Without them it will be difficult to troubleshoot your work.
 
A dad pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 18 rpm in 13.0 . Assume the merry-go-round is a uniform disk of radius 3.0m and has a mass of 760 kg, and two children (each with a mass of 21kg ) sit opposite each other on the edge.

Calculate the torque required to produce the acceleration, neglecting frictional torque.

What force is required at the edge?



2. Homework Equations

torque= I alpha
w=2pi f
w=w0+alpha *t

3. The Attempt at a Solution

change f to w
w= 2pi *.3rev/sec w=1.8849 rad/sec
find alpha
w=w0+ alpha*t 1.8849=alpha*13 alpha = .144997rad/s^2

find I
I= .5 MR^2= 3609 kg m^2
torque= I alpha = 3609*.144997= 523 N M this answer was incorrect I do not know why I also tried it with I for a hoop the answer was 1046 N M but this answer was also incorrect if you can tell me where i messed but that would be very helpful thanks

The question is what torque is required for this acceleration and what force is required at the edge?
 
Consider the children. They add to the moment of inertia of the system.

Forget the hoop. The problem specifically says that the merry-go-round should be considered as a disk.
 
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