- #1

zizzle

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- 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.

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.