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

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In summary: Then the moment of inertia of the system will be the sum of the moment of inertia of the children and the moment of inertia of the disk.The moment of inertia of the disk is given by I = 1/2 MR^2, where M is the mass and R is the radius. In this case, M = 760 kg and R = 3.0 m. So I = 1/2 * 760 * (3.0)^2 = 3420 kg m^2.The moment of inertia of the children can be calculated by treating them as point masses at a distance of 3.0 m from the center of the disk (since they are sitting on the edge).So
  • #1
kaite
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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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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 

What is "Merry go round torque"?

"Merry go round torque" is the force that causes a rotating object, such as a merry go round, to accelerate. It is the product of the force applied to the object and the distance from the axis of rotation.

How is "Merry go round torque" calculated?

To calculate "Merry go round torque", you need to know the force applied to the object and the distance from the axis of rotation. The formula for torque is T = F x d, where T is torque, F is force, and d is the distance from the axis of rotation.

What factors affect "Merry go round torque"?

The two main factors that affect "Merry go round torque" are the force applied to the object and the distance from the axis of rotation. In addition, the mass and distribution of the object, as well as the speed of rotation, can also impact the torque.

Why is "Merry go round torque" important to understand?

Understanding "Merry go round torque" is important because it helps us to predict and control the motion of rotating objects. It is also crucial in designing and operating machines, such as engines and turbines, that rely on rotational motion.

How is "Merry go round torque" related to angular momentum?

"Merry go round torque" is directly related to angular momentum, as it is the cause of changes in an object's rotational motion. The greater the torque applied, the greater the change in angular momentum. This relationship is described by the equation T = ΔL/Δt, where T is torque, ΔL is change in angular momentum, and Δt is the time interval over which the torque is applied.

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