In a “Rotor-ride” at a carnival, people rotate in a vertical c...

Question

In a “Rotor-ride” at a carnival, people rotate in a vertical cylindrically walled “room.” See Fig. 5–52. If the room radius is 5.5 m, and the rotation frequency 0.50 revolutions per second when the floor drops out, what minimum coefficient of static friction keeps the people from slipping down? People on this ride said they were “pressed against the wall.” Is there really an outward force pressing them against the wall? If so, what is its source? If not, what is the proper description of their situation (besides nausea)? [Hint: Draw a free-body diagram for a person.]
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Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A washing machine operates its spin cycle to remove water from clothes. The drum of the washing machine has a radius of 0.30 m and rotates at a frequency of 1600 revolutions per minute R PM. Determine the minimum coefficient of static friction between the clothes and the inner surface of the drum required to prevent the clothes from slipping down as the drum spins. So that's our angle is where ultimately we're trying to figure out what the minimum coefficient of static friction value must be in order to keep the clothes from slipping down the inner surface to the bottom of the washing machine drum as it spins. Awesome. And that will be our final answer that we're trying to solve for is this minimum value for the coefficient of static friction. So looking at our picture that's provided to us, we have a cylinder shape that denotes our drum, our washing machine drum to be exact. And it will, it appears to be rotating to the right in a clockwise motion. And we have a dotted blue line going through the center of our drum to the bottom. So from the top to the bottom and, and from this midline, the center line from the center of our drum to the outer edge of the drum has some value of radius R which is given to us as 0.30 m. And we have our clothes that are inside of this washing machine drum as it's spinning around with their green arrows denote the spinning. Awesome. So with that in mind now that we have a visualization of what's going on in this problem, let's look at our multiple choice answers to see what our final answer might be. A is 1.2 multiplied by 10 to the power negative three. B is 2.0 multiplied by 10 to the power negative three C is 2.5 multiplied by 10 to the power of negative three and D is 5.0 multiplied by 10 to the power of negative four. OK. So first off, let us make the following assumptions. Firstly, let us assume other forces such as air resistance are inedible, meaning we can ignore it. And secondly, let us assume that the interaction between the clothes and the drum's inner surface is simplified to a flat contact. Let us also note that the drum's rotation will cause the clothes to press against the drum's inner wall surface and will generate a normal force that is perpendicular to the surface. Let us also note that the static frictional force which acts parallel to the surface of the drum must be of a sufficient value. In order to counteract the weight of the clothes, let us draw a more detailed diagram. Now to help us better visualize all the forces at work in this problem. So in this diagram that we have here, we have a pink and blue blob representing the clothes and our downwards red arrow is denoting the gravitational force. And the upwards arrow red arrow is denoting the frictional force and perpendicular to the clo from the clothes on the wall. Going towards the right side of the washing machine drum is the normal force. Awesome. So now and then also the gray cylinder here, just a little glimpse of just a piece of the drum instead of the whole drum. That's what the gray area denotes is the inside of the drum, the washing machine drum. So now at this point, we must solve for the angular velocity using the rotational rate. So let us recall and use the angular velocity equation which states that Omega which omega denotes the angular velocity and the angular velocity is equal to two multiplied by pi multiplied by F where F is the frequency. So now we could plug in all of our known values to solve for the angular velocity. So the angular velocity omega is equal to two multiplied by pi multiplied by the frequency which is given to us as 1600 revolutions per minute. So RP MS but we have a bit of a problem, we need to convert RP MS to radiance per second. So we need to use dimensional analysis. So we need to multiply by in one minute, there is 60 seconds. Awesome. So that will mean when we simplify that Omega which is our angular velocity. So our angular velocity omega is equal to two multiplied by pi multiplied by 1600 divided by 60. And its units are now in radiance per second. Awesome. So now at this point, we must solve for the tangential or the linear velocity by recalling and using the following equation that V the velocity is equal to the angular velocity multiplied by R which is equal to when we plug in our known variables, two multiplied by pi multiplied by 1600 divided by 60 radians per second, multiplied by hour radius value which is 0.30 m is RTA radius, which when we plug that into our calculator V will be equal to 50.2655 when we round to four decimal places. And don't forget that its units are meters per second. So now at this stage, we need to solve for the net horizontal force by recalling and using the net vertical, sorry. So now we need to solve for the net horizontal force. I was getting ahead of myself. So to solve for the net horizontal force, we need to recall and use the following equation that the normal force FN is equal to FC, which is equal to M multiplied by V squared divided by R. And now now we could start solving for the net vertical force. So to solve for the net vertical force, we need to recall the following equation that the frictional force F subscript F is equal to the gravitational force which is equal to zero. So we could write, we could take it even further and write that the frictional force is equal to the gravitational force, which we can take it even further and write that the static friction, which is denoted as mu subscript S multiplied by the normal force which is F subscript N is equal to mass multiplied by gravity. Awesome. So now at this point, we need to rearrange this equation to isolate and solve for us, which is our final answer. So us multiplied by M multiplied by V squared divided by R. Is that our or centripetal force equation equals M multiplied by G which we can simplify once we could keep simplifying to get that us multiplied by V squared because you'll see that the M will cancel out. So V squared divided by R is equal to G So then we can simplify once again and finally get MU S by itself. And mu S is equal to R multiplied by G divided by V squared, which is equal to when we plug in our known values, 0.30 m which is R radius multiplied by G which G denotes the acceleration due to gravity, which is 9.8 m per second squared. That's the numerical value for the acceleration due to gravity divided by V squared, which we found our V value to be 50.2655 m per second. And this value is squared. So when we plug that into our calculator, we will get 0.001116 which when we write that in scientific notation, because all of our multiple choice answers are in scientific notation, we will get 1.2 when we round to one decimal place, multiplied by 10 to the power of three. And that is our final answer. Hooray, we did it and this corresponds with multiple choice answer letter A 1.2 multiplied by 10 to the power of negative three. And that's it we solve for this problem. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

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