In an old-fashioned amusement park ride, passengers stand inside a 4.9-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick" to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.60 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed." What is the minimum angular speed, in , for which the ride is safe? I have no idea how to go about this problem. I know angular velocity is w = v/r but i just dont know how to apply it to get the answer to this question. someone help please!!
Welcome to Physics Forums! When the cylinder is rotating and the riders are stationary against the wall, what can you say about the forces in the vertical direction? Moderation note: I've moved this from Adv. Phys. to Intro. Phys.
Welcome to PF! Hi lmc489! Welcome to PF! Hint: use centripetal acceleration EDIT: oooh! Hi Hoot! You beat me to it!
... if the riders aren't moving up or down, then the only forces that matter are in the x direction, right? the forces in the y direction must cancel. (well, not move, but accelerate, i guess) is the force that is moving the people in the ride the friction they have against the wall?
thoughts: if the person is sticking to the wall, then the net force in the y direction is zero. so what must be moving the person around the circle is the friction between their clothing and the steel wall behind them. this has to be static friction... because otherwise, the person would be sliding all over the place. so... Fnet = fs = UsN (where Us is the static coefficient of friction, and N is the normal force) because i'm trying to find the slowest angular speed, i want to use the lower of the two coefficients of static friction. and N is going to be equal to mg, or 30 kg (the lightest a kid on the ride can be) times acceleration due to gravity. this force of friction will have to equal (mv^2)/r and then the v there can be turned into angular velocity using the w= v/r equation (and then converted to proper units using dimensional analysis...) does that make sense?
No this does not make sense, but you're on the right lines. The frictional force acts vertically upwards, the centripetal forces acts, by definition towards the centre of rotation. Therefore, the frictional force is perpendicular to the centripetal force! So how can the frictional force be responsible for the centripetal force?
Welcome to PF! Hi hddock! Welcome to PF! But please don't give such extensive answers (even if they're right) … on this forum, we aim to give hints, so that OPs do the problems themselves … if the OP is given too much detail, they won't have had enough practice to do it in the exam … and please don't give extra help when the OP hasn't had the chance to react to the previous help. The OP may thank you, and may get extra marks for it, but it's cheating, and it won't help in the exam. Your help would have been fine if you had stopped at your first answer. Look at the style of Hootenanny's reply and mine, and help by pushing rather than dragging.