A father spins his child on a cart

In summary, the father is standing on a conical hill and spinning his child on a cart using a 2.0-m-long rope. The rope tension needed for the cart to spin at 14 rpm is 178.7N, taking into account the horizontal plane of rotation and the upslope component of the force. The normal force and tension both contribute to the centripetal force.
  • #1
Cc518
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Homework Statement


The father stands at the summit of a conical
hill as he spins his 20 kg child around on a 5.0 kg cart with a
2.0-m-long rope. The sides of the hill are inclined at 20 degrees. He
keeps the rope parallel to the ground, and friction is negligible.
What rope tension will allow the cart to spin with the same
14 rpm it had in the example?

Homework Equations


F=mrw^2

The Attempt at a Solution


I drew the fbd diagram, and I got two equation :1. Fny +Ty=mg and 2. Tx-Fnx=mrw^2
where Fnx and Fny is the x, y-component of the normal force, Tx and Tx is the x, y-component of the Tension.
But I don't know how to solve for Fn
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Can anyone help me?
The link to the image: https://cloud.smartdraw.com/share.aspx/?pubDocShare=3584F28114431EA1F27BA700C6061213131
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  • #2
You have two unknowns, Fn and T. You want to find T, not Fn. use one equation to eliminate Fn from the other.
 
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  • #4
Cc518 said:
Thank you for reply!
I got tention as 179.2n, but the answer I got from other sources is 185n.
The fbd they used is different than mine though.
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https://cloud.smartdraw.com/share.aspx/?pubDocShare=3584F28114431EA1F27BA700C6061213131
They said Tention= mrw^2-mgx
Which one is correct?
I get 178.7N.
Did you mean they get mrω2+mgx?
That overlooks that the rotation is in the horizontal plane, so need to take the upslope component of mrω2 to align it with the tension. And r here, of course, is the radius of the rotation, not the length of the rope.
 
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  • #5
haruspex said:
I get 178.7N.
Did you mean they get mrω2+mgx?
That overlooks that the rotation is in the horizontal plane, so need to take the upslope component of mrω2 to align it with the tension. And r here, of course, is the radius of the rotation, not the length of the rope.

Sorry, I meant mrω2+mgx actually.
So is the rotational plane horizontal or is it parallel with the incline of the slope?
In other words, does the Tx act as centripetal force or does T act as the centripetal force?
Thank you :)
 
  • #6
Cc518 said:
is the rotational plane horizontal or is it parallel with the incline of the slope?
What do you think?
 
  • #7
haruspex said:
What do you think?
I think it should be horizontal because if the plane is parallel to the incline of the slope, the object, in this case, the cart, will move up the slope, which is not what happened.
 
  • #8
Cc518 said:
I think it should be horizontal because if the plane is parallel to the incline of the slope, the object, in this case, the cart, will move up the slope, which is not what happened.
Right.
 
  • #9
haruspex said:
Right.
Thank you so much for your help! I really appreciated it!
 
  • #10
Cc518 said:
does the Tx act as centripetal force
It contributes to the centripetal force. The normal force also affects it.
 
  • #11
haruspex said:
It contributes to the centripetal force. The normal force also affects it.
Oh, yeah. Thank you for the reminder :)
 

What is the physics behind a father spinning his child on a cart?

The physics behind a father spinning his child on a cart involves the principle of angular momentum. When the child is rotating on the cart, they have a certain amount of angular momentum, which is conserved throughout the motion. The father's actions of pulling the cart and spinning it in a circular motion increases the child's angular momentum. This results in the child moving in a circular path, as well as experiencing a centrifugal force outward from the center of rotation.

Why does the child feel dizzy after being spun on a cart?

The feeling of dizziness after being spun on a cart is due to the child's vestibular system. This system is responsible for maintaining balance and orientation in the body. When the child is spun on the cart, their vestibular system is disrupted, causing them to feel dizzy and disoriented. This feeling usually goes away once the child stops spinning and their vestibular system has a chance to readjust.

What are the potential dangers of spinning a child on a cart?

The potential dangers of spinning a child on a cart include the risk of the child falling off the cart and getting injured. The centrifugal force created by the spinning motion can also cause the child to lose their balance and fall off. Additionally, spinning the child too fast or for too long can also lead to feelings of nausea and dizziness, which can be uncomfortable for the child.

Can spinning a child on a cart cause any long-term effects?

There is no evidence to suggest that spinning a child on a cart can cause any long-term effects. However, it is important to ensure that the child is not being spun too fast or for too long, as this can lead to feelings of discomfort and nausea. It is also important to monitor the child's safety while spinning to prevent any potential injuries.

Are there any educational benefits to spinning a child on a cart?

Spinning a child on a cart can have some educational benefits, such as teaching them about the concept of angular momentum and centrifugal force. It can also be a fun way to introduce them to physics and encourage their curiosity about the world around them. However, it is important to ensure that the child's safety is the top priority and that they are not spun too fast or for too long.

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