Your roommate is working on his bicycle and has the bike upside down. He spins the 60-cm-diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. What are the pebble's speed and acceleration?

Video transcript

Hey, everyone in this problem. A wheel has a decoration mounted 20 centimeters from the axis of the 700 millimeter wheel. An observer notices that the wheel makes five turns every second. And we're asked to calculate the speed in meters per second and the net acceleration of the decoration. We're given six answer choices, options A through F each of them containing a different combination of speed and acceleration. Now, starting with this speed, OK. We're asked to find this in meters per second. What that means is that we're looking for that tangential speed, not the angular speed. OK. All right. So how can we find the tangential speed? I'll recall that the tangential speed V T is equal to R multiplied by Omega. OK. R is the radius or that distance from the axis and Omega is the angular speed. So we know our, and the distance from the decoration to the axis is 20 centimeters. That's gonna be our, our value in Omega. We're given some information about Omega when we're told that this makes five turns every second. So Omega is going to be equal to five revolutions per second. And we want to convert this to our standard unit of radiance per second. So we're gonna multiply and we know that there are two pi radiant in every revolution. And you can think about going around a circle one complete time. That's two pi radiant. The unit of revolution divides out and we're left with the omega which is 10 pie radiance per se. So we have omega and what about R we've said that R is gonna be centimeters. Now, our answer choices are in meters per second. So we want to convert this into meters. So we multiply by one m divided by 100 centimeters. OK? Because we know there are 100 centimeters in every meter, the unit of centimeter divides out and we're effectively dividing by 100 to get 0.2 m. So now we have our R, we have our Omega and we can calculate this tangential speed that we're looking for V T. It's gonna be equal to 0.2 m multiplied by 10 pi radiance per second. A dotted line there to separate, working out our units versus working on our tangential velocity. OK. This is gonna give us a speed of two pi meters per second and we can simplify and we're gonna round to two decimal places and this gives us 6.28 m per second. OK? So we have our speed down 6.28 m per second. If we look at our answer choices options. B ce and F have a different speed than 6.28 m per second. Ok. So we're gonna eliminate all of those options and we're gonna be left with option A or D and now they have the speed of 6.28 m per second that we just found they have different accelerations. So let's move to the second part of this problem and calculate that acceleration. And we can see that the answer choices give the acceleration A R and that is in meters per second squared. So what this tells us is that we're looking for that radial acceleration of the centripetal acceleration. OK. So let's recall that that centripetal acceleration is given by V squared divided by R. Oh we know V now we just calculated V and we know R so this is a simple problem of substituting in these values. So for V let's go back to two pi meters per second and that's before we head round it. So let's use that value. So we avoid any round off error. We get two pi meters per second, all squared. And that's gonna be divided by R which is 0. m. This is gonna give us an acceleration a of 197.4 m per second squared. And we had meters squared per second squared, we divided by meters and we get that unit for acceleration that we were looking for. So now we have our speed 6.28 m per second, our acceleration of 197.4 m per second squared and this is gonna correspond with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.