In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in
FIGURE P8.51. b. What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

Question

In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in 
FIGURE P8.51. b. What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

Accepted Answer

Hi, everyone. In this practice problem, we are being asked to find the maximum rotation period of a drum of a washing machine. So we'll have a washing machine with a 40 centimeter diameter rotating about a horizontal axis. A child actually put her 300 g doll and set up the washer. And we're being asked to find the maximum rotation period that will keep the doll stuck to the drum at its highest position. The options given are a 0.9 seconds. B 1.2 seconds, C 1.6 seconds and D 5.0 seconds. So in this particular practice problem, the doll will undergo a vertical circular motion at the top of its motion. The doll will be subjected to two different forces which I will indicate in our diagram right here. So we'll have this circle here which is going to be the trajectory of the vertical circular motion. I'm going to represent the doll with the sphere right here. And we are being asked to find a maximum rotation period of the drum that will keep the doll stuck to the drum at its highest position. So this is going to be the highest position of the doll inside of the washing machine itself. So as I said, the doll will experience two different forces. Uh The first one is going to be the weight of the doll itself, which is going to be represented by this red arrow right here, which is going to be M multiplied by G. The second one is going to actually be the normal force as a result of the doll um sticking onto the door, the wall of the washer drum. So the normal force is going to also be pointing downwards like so just like the weight and I'm going to represent that with an end. So both the weight and also the normal force will be ver uh pointing vertically downward. And now, what we wanna do is to actually apply Newton's second law in order for us to find the limiting velocity and therefore the maximum rotation period. All right. So according to Newton's second law, the sigma of all the forces acting on our system in the radial direction, which in this case, our system is going to be the dull will equals to the mass multiplied by the acceleration in the radial direction as well. The total of the two forces are going to be M G plus N, which is the normal force, which will equals to M multiplied by A R. We wanna recall that A R equals to V squared divided by the radius of the trajectory itself, the V is going to be the linear velocity. So we wanna substitute this A R into our new and second law in order to give us M G plus N equals M multiplied by V squared divided by R just like. So next, what we wanna do is to Ashley um note that at the instant when the normal force becomes zero, then the doll will actually fall. So what we wanna do is we want to determine the limiting velocity or the limit speed at the top or at the highest position that will keep the doll stuck to the drum. So at normal Um equals to zero new 10, I'm gonna write Newton just so that we are not getting confused at the normal force equals to zero Newton or when it is equals to zero Newton, then the velocity is going to be the limiting velocity that will keep the doll stuck to the drum. And we want to substitute this known condition into our Newton second law expression. So M G plus N equals M multiplied by V squared divided by R when substitute the zero Newton. So M G will equals to M multiplied by V limit squared divided by R. We can actually uh cross out the masses into here. So that will give us um equation four G equals V limit squared divided by R just like. So, so next, we have the equation essentially for the V limit through just some manipulation of the equation here. But we wanna notice that the V or the velocity have a relationship with the rotational period, which is what we are interested at. So the relationship between linear speed V and the rotation period is given by this following equation right here. So V will equals to R multiplied by omega. And we want to recall that Omega will equals to two pi divided by T, which is the period. So we want to substitute that. So V equals R multiplied by two pi divided by T. And we can actually utilize this equation right here to change it with V limit. So when the velocity as when the velocity equals to P limit, then the T will equals to T max. So that will give us a equation of V limit. Well equals to R multiplied by two pi divided by T max just like. So, and this T max is the one that we are interested at. Awesome. So we wanna combine this two equation right here in order for us to solve what the T max is. So we get G equals V limit squared divided by R and that equals to uh R Multiplied by two pi divided by T max squared multiply by one over R just like. So, so I will essentially just um take the T X and put it on to the other side so that we get uh an expression for T max. So you essentially just wanna manipulate this equation right here in order for us to get the equation for T max. So T X, well, then Ashley equals two, two pie multiplied by the square root of R divided by G. And this is the final equation that we have considering that we manipulated the previous expression. So this will give us two pi multiplied by the square root of the R which is going to be 0.20 m or 0.20 m, Which come from the fact that the r equals 20 cm and G will still equals to 9.81 m per second squared. That will give us a T max value of 0.90 seconds. And that will actually be the answer to this particular practice problem. So the maximum rotation period of the drum that will keep the doll stuck to the drum at its highest position is going to be equals to 0.9 seconds. And that will correspond to option a in our answer choices. So answer a is going to be the answer to this particular practice problem. And that will be all for this one. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topic and that'll be all for this one. Thank you.

Verified Solution

Key Concepts

Centripetal Force

Gravitational Force

Period of Rotation