Minimum initial speed to spin a particle around a disk (with gravity)

[zizzle]

Homework Statement

A popular ride at some playgrounds consists of two circular disks joined by rungs (Figure 1). The center of the unit is attached to a horizontal axle around which the unit rotates. The rotational inertia of the unit is 250 kg⋅m^2 , and its radius is 1.00 m. A 30-kg child runs and grabs the bottom rung of the ride when it is initially at rest. Because the child can pull herself up to meet the rung at waist height, treat her as a particle located on the rung. The friction between unit and axle is minimal and can be ignored.

a). What is the minimum speed v the child must have when she grabs the bottom rung in order to make the unit rotate about its axle and carry her over the top of the unit?

b). Is this a reasonable speed for a child?

Relevant Equations

Rotational KE=1/2*I*w^2
v=wr

For this problem, since the weight force on the "particle" (child) is not always aligned with the tangential circular path of the disks, I couldn't think of a way to use rotational kinematics equations.

As such, I tried to solve the problem using work principles (namely, that the change in energy of the system is the work done). Since the centripetal force on the particle is perpendicular to the motion, work done is zero, so I said that:
1/2*I*w^2 = mgh
aka initial rotational velocity equals final gravitational potential energy at a height of 2 (when the particle has made it to the top of the unit).
Solving for v (initial speed):
v=r*√(2mgh/I).
I checked the units for this equation, and they checked out to be in m/s, so I thought I had found the answer.

I initially tried values of m=30kg, g=9.81 m/s^2, h=2 (assuming bottom of unit is 0), and I=250 kg m^2. This was not correct. I then realized that the 'I' of the system would change with the child on board. So I said 250=mr^2, since r is 1, then the mass of the disk should be 250 kg. So the moment of inertia of the system including the child would be I=250+30(1)^2=280. Using this value was not correct, either.
Also, my answers for these two were around 2.5 m/s, but I've determined that the answer to (b) is "This is not a reasonable speed for the child", but 2.5 m/s isn't too unreasonable. So it's obvious that's not correct.

Any hints or help would be greatly appreciated.

 

[PeroK]

Homework Statement: A popular ride at some playgrounds consists of two circular disks joined by rungs (Figure 1). The center of the unit is attached to a horizontal axle around which the unit rotates. The rotational inertia of the unit is 250 kg⋅m^2 , and its radius is 1.00 m. A 30-kg child runs and grabs the bottom rung of the ride when it is initially at rest. Because the child can pull herself up to meet the rung at waist height, treat her as a particle located on the rung. The friction between unit and axle is minimal and can be ignored.

a). What is the minimum speed v the child must have when she grabs the bottom rung in order to make the unit rotate about its axle and carry her over the top of the unit?

b). Is this a reasonable speed for a child?
Homework Equations: Rotational KE=1/2*I*w^2
v=wr

For this problem, since the weight force on the "particle" (child) is not always aligned with the tangential circular path of the disks, I couldn't think of a way to use rotational kinematics equations.

As such, I tried to solve the problem using work principles (namely, that the change in energy of the system is the work done). Since the centripetal force on the particle is perpendicular to the motion, work done is zero, so I said that:
1/2*I*w^2 = mgh
aka initial rotational velocity equals final gravitational potential energy at a height of 2 (when the particle has made it to the top of the unit).
Solving for v (initial speed):
v=r*√(2mgh/I).
I checked the units for this equation, and they checked out to be in m/s, so I thought I had found the answer.

I initially tried values of m=30kg, g=9.81 m/s^2, h=2 (assuming bottom of unit is 0), and I=250 kg m^2. This was not correct. I then realized that the 'I' of the system would change with the child on board. So I said 250=mr^2, since r is 1, then the mass of the disk should be 250 kg. So the moment of inertia of the system including the child would be I=250+30(1)^2=280. Using this value was not correct, either.
Also, my answers for these two were around 2.5 m/s, but I've determined that the answer to (b) is "This is not a reasonable speed for the child", but 2.5 m/s isn't too unreasonable. So it's obvious that's not correct.

Any hints or help would be greatly appreciated.

If you came along and gave the bottom rung a push to get the wheel moving what would it do after that?

 

[haruspex]

initial rotational velocity equals final gravitational potential energy

I assume you mean initial rotational energy.

solve the problem using work principles

You can, but only once the initial rotational energy has been found.
Do not assume the process of grabbing the rung is work conserving. How else can you figure out the initial rotational speed?

 

[zizzle]

If you came along and gave the bottom rung a push to get the wheel moving what would it do after that?

Assuming no friction, it'd keep moving in a circle forever due to conservation of angular momentum, would that tie into anything?

I assume you mean initial rotational energy.

You can, but only once the initial rotational energy has been found.
Do not assume the process of grabbing the rung is work conserving. How else can you figure out the initial rotational speed?

Would I have to use rotational kinematics? I can't seem to find a way to do that since the angular acceleration wouldn't be constant, right? (due to the weight force of the child)

 

[PeroK]

Assuming no friction, it'd keep moving in a circle forever due to conservation of angular momentum, would that tie into anything?

It might make the calculations simpler. To what extent is the wheel irrelevant?

Would I have to use rotational kinematics? I can't seem to find a way to do that since the angular acceleration wouldn't be constant, right? (due to the weight force of the child)

Hint: think collision.

 

[haruspex]

Would I have to use rotational kinematics?

Not kinematics, no.
Here's a general tip for all motion problems: think through the conservation laws and consider which should apply.

 

[zizzle]

Not kinematics, no.
Here's a general tip for all motion problems: think through the conservation laws and consider which should apply.

So, I got the correct answer, but I still have some questions.

I first realized that angular momentum would be conserved (even though the energy wasn't conserved until after the hit). So L_i=mv_ir, and L_f=Iw_f+v_fmr.
Thus mv_ir=Iw_f+v_fmr.
Solving for v_i:

v_i=(Iw_f+v_fmr)/mr=(Iw_f/mr)+v_f=(Iv_f/mr²)+v_f=v_f(I/mr²+1)

At this point, I was pretty lost, so I contacted a TA in my class for help. They briefly mentioned using a conservation of energy equation in order to find v_f. They gave me this equation:

mgh=(1/2)Iw_f²+(1/2)mv_f²

I turned the w_f into v_f/r and solved for v_f. Then I plugged it into the equation from before and got a correct answer.

However, my question is, why would mgh=(1/2)Iw_f²+(1/2)mv_f² hold? Shouldn't the final potential energy equal the initial kinetic energy (rather than the final KE?)

 

[haruspex]

Shouldn't the final potential energy equal the initial kinetic energy (rather than the final KE?)

It depends what you mean by "final velocity " here.
There are two steps in the sequence. First, from just before the boy grabs a rung to just after; second, from then until the highest point of the rotation. There is an initial velocity (or angular velocity) and a final such for each step.
Yes, the correct equation is mgh=(1/2)Iw_i²+(1/2)mv_i², but that is where v_i etc. mean the initial values for the second step, which is to say the final values from the first step.

 

[zizzle]

It depends what you mean by "final velocity " here.
There are two steps in the sequence. First, from just before the boy grabs a rung to just after; second, from then until the highest point of the rotation. There is an initial velocity (or angular velocity) and a final such for each step.
Yes, the correct equation is mgh=(1/2)Iw_i²+(1/2)mv_i², but that is where v_i etc. mean the initial values for the second step, which is to say the final values from the first step.

Ah, I understand now! I'm not sure why that was difficult for me to wrap my head around, since I've dealt with "split problems" just like this one in linear motion scenarios. As soon as rotational motion comes into play, my brain seems to shut off. Thanks so much for the help :)

 

[PeroK]

Ah, I understand now! I'm not sure why that was difficult for me to wrap my head around, since I've dealt with "split problems" just like this one in linear motion scenarios. As soon as rotational motion comes into play, my brain seems to shut off. Thanks so much for the help :)

If a problem is more complicated, I prefer to number things. In this case , I would have:

$E_0 = \frac 1 2 mv_0^2$

As the initial energy and speed of the child.

$E_1 = \frac 12 mv_1^2 + \frac 12 I \omega_1^2$

As the energy of the system after the "collision".

And

$E_2 = mgh = mg(2R)$

as the (final) energy of the system - assuming the wheel just comes to rest as the child reaches the top.

It feels a lot cleaner to me to set the problem up like this. Once you have $v_f$ as an intermediate velocity it feels like the notation is started to lead you astray.

It also allows you to express ideas more easily than were able to. For example:

"By conservation of energy during the rotation we have $E_1 = E_2$."

Makes it easier to see what you are doing.

Likewise, I'd have $L_0$ and $L_1$ for angular momentum about the centre of the wheel. And $L_2 =0$.