# Physics merry-go-round problem

• Jamest39
In summary: The force would just be the acceleration of the merry-go-round? which would be a=v^2/r or ω^2 * rFor a heavy package a greater force is required than for a light package.
Jamest39

## Homework Statement

A small package is placed on the platform of a non-rotating merry-go-round, 2.72 m from the vertical rotation axis. At t=0, the merry go round begins rotating with a constant angular acceleration of α = 0.100 rad/s^2. The coefficient of static friction is 0.600. Determine the time t at which the package first begins to slide.

ω = ωinital + αt

## The Attempt at a Solution

My thinking of it was that the package would begin to slide after the max of its static friction is reached (μN), but was having trouble applying that and also expressing my answer in terms of time that has passed.

berkeman
Hello James,

So until it slides it makes a circular motion. What force is required to keep it in this circular trajectory ? (Magnitude in terms of ##\omega##?)

Yes you are correct that the object will begin to slide once the maximum of the static friction is reached, but what force can you think of that will cause the object to start sliding?

BvU said:
Hello James,

So until it slides it makes a circular motion. What force is required to keep it in this circular trajectory ? (Magnitude in terms of ##\omega##?)

The centripetal acceleration will be what is keeping it in it's circular trajectory: a = v^2/r = ω^2 r, so in terms of omega: ω = sqrt(a/r)

Good. What is providing the force that results in that centripetal acceleration ?

BvU said:
Good. What is providing the force that results in that centripetal acceleration ?

Would it the spinning of the merry-go-round with the constant angular acceleration?

Like: the more it spins the more centripetal force it provides ?

BvU said:
Like: the more it spins the more centripetal force it provides ?

I think it would be the faster it spins the more centripetal force it will provide

Suppose you would be sitting there and it would be a bit slippery...

BvU said:
Suppose you would be sitting there and it would be a bit slippery...

After it starts constantly accelerating, eventually there will be a point where the velocity is high enough to overcome the friction and cause me to slide around on it

Yes, in a way. It is the friction that keeps you in a circular trajectory. No friction and you start sliding outwards straightaway.

So you have an expression for the force required to keep the package in a circular trajectory, and you have an expression for the maximum force the friction can provide...

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BvU said:
Yes, in a way. It is the friction that keeps you in a circular trajectory. No friction and you start sliding outwards straightaway.

So you have an expression for the force required to keep the pacakage in a circular trajectory, and you have an expression for the maximum force the friction can provide...

Ah, I see. So the centripetal acceleration is keeping the merry-go-round in motion, but it is the friction that is keeping the package in motion and not moving.
And for the circular trajectory, its ω = sqrt(a/r) and the maximum friction in μ * N (normal force).

Jamest39 said:
So the centripetal acceleration is keeping the merry-go-round in motion
Oops, no, there is some electric motor underneath that provides the torque for uniformly accelerated angular motion. The merry-go-round itself is a solid construction that stays together and thus provides its own centripetal force.
The friction is keeping the package moving on a circular trajectory !

It only appears not to be moving if you are on the merry go round: it does not move with respect to the merry-go-round.

Can you manipulate your expression ##\omega = \sqrt {a/r} ## in such a way that you get the expression for the force required to keep the package in its circular trajectory ?

BvU said:
Oops, no, there is some electric motor underneath that provides the torque for uniformly accelerated angular motion. The merry-go-round itself is a solid construction that stays together and thus provides its own centripetal force.
The friction is keeping the package moving on a circular trajectory !

It only appears not to be moving if you are on the merry go round: it does not move with respect to the merry-go-round.

Can you manipulate your expression ##\omega = \sqrt {a/r} ## in such a way that you get the expression for the force required to keep the package in its circular trajectory ?

The force would just be the acceleration of the merry-go-round? which would be a=v^2/r or ω^2 * r

For a heavy package a greater force is required than for a light package. I don't see that in ##\omega^2 r##

BvU said:
For a heavy package a greater force is required than for a light package. I don't see that in ##\omega^2 r##

Haha, I guess not. I mean, would it have something to do with other forces acting on it? Like gravity (mg) and the normal force from the ground of the merry-go-round (N)?

No, what I mean is the ##\omega^2r## has the wrong dimension. It is an acceleration. According to good old Isaac, Force = mass times acceleration. So you need a factor ##m## in there !

BvU said:
No, what I mean is the ##\omega^2r## has the wrong dimension. It is an acceleration. According to good old Isaac, Force = mass times acceleration. So you need a factor ##m## in there !

Ohhh, obviously. So something like ∑F = m * ω^2 * r

Bingo. This is the force that is needed to keep the package in a circular trajectory. And ##\omega## is known as a function of time ##t##.

The only force that can do that is the friction force with the floor, and you already had an expression for the maximum friction force ##F_{\rm fric, \ max} = \mu N## (to be worked out in known variables -- again something with the mass ##m## of the package )

As soon as required force > max friction the package can no longer be kept in a circular orbit !

BvU said:
Bingo. This is the force that is needed to keep the package in a circular trajectory. And ##\omega## is known as a function of time ##t##.

The only force that can do that is the friction force with the floor, and you already had an expression for the maximum friction force ##F_{\rm fric, \ max} = \mu N## (to be worked out in known variables -- again something with the mass ##m## of the package )

As soon as required force > max friction the package can no longer be kept in a circular orbit !

So I would need to calculate the time at which that maximum friction equals the force?: m*ω^2*r = μ N
If so, if so, how can I go about finding the angular velocity from the know angular acceleration? Also the normal force that's being exerted on the package?

Normal force is as always: ##mg##.
And your relevant equation in post #1 gives you ##\omega(t)## !
Piece of cake !

BvU said:
Normal force is as always: ##mg##.
And your relevant equation in post #1 gives you ##\omega(t)## !
Piece of cake !

So, using that equation, I would take m * (ωintial + αt)^2 * r = μmg and then just solve this for t? And since its starting from rest, the ωinitial is just zero, right?

Excellent ! And, as you see, the mass divides out conveniently.

Jamest39
BvU said:
Excellent ! And, as you see, the mass divides out conveniently.

Awesome! Thank you so much! The answer to this problem was provided to us (we have to show our work of course), and I just did the calculations and it is right!

A good achievement! The take-home message is that the term centrifugal force is somewhat misleading. It is a fictitious force you experience when you are in a merry-go-round.

Newtons law says that in the absence of any force things go straight with the inital velocity they happen to have.

For a circular trajectory a centripetal force is required. It can be provided by a wire, by holding onto something, by friction or even by gravity (road turns that lean sideways).

Jamest39
Aren't you missing something ?
In the reference frame of the package, it has two perpendicular components of acceleration, one arising because our frame is non-inertial, i.e. the centrifugal force AND the acceleration arising from the angular acceleration of the merry go round. And thus the friction would be the resultant of these, one component providing the centripetal acceleration and the other against the tangential acceleration.

haruspex
The exercise wants you to ignore the angular acceleration in this stage of the curriculum. The unrealistically low value of 0.1 rad/s2 indicates so (and the book answer confirms). But mind is correct.

 Nonsensical statement of the week ! Been to the fair and a merry-go-round doesn't go much faster than one revolution per 15 seconds, 0.4 radians per second. The 0.1 rad/s2 isn't unrealistic, but I do have my doubts the exercise result for ##t## is. We don't want little kids to fly off the thing, do we ?

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## 1. What is the Physics merry-go-round problem?

The Physics merry-go-round problem, also known as the "spinning disk problem", is a classic physics problem that involves calculating the angular velocity and acceleration of a rotating disk or merry-go-round.

## 2. What are the key components of the Physics merry-go-round problem?

The key components of the Physics merry-go-round problem include the radius of the rotating disk, its mass, and the force applied to the disk, as well as the angular velocity and acceleration of the disk.

## 3. How is the Physics merry-go-round problem solved?

The Physics merry-go-round problem is typically solved using Newton's second law of motion, which states that the sum of all forces acting on an object is equal to its mass times its acceleration.

## 4. What are the main applications of the Physics merry-go-round problem?

The Physics merry-go-round problem has many real-world applications, such as understanding the motion of planets and satellites, analyzing the behavior of rotating machinery, and designing amusement park rides.

## 5. What are some common misconceptions about the Physics merry-go-round problem?

One common misconception about the Physics merry-go-round problem is that the angular velocity of the disk is constant. In reality, the angular velocity changes as the disk accelerates or decelerates due to the applied force.

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