A Ferris wheel of radius R speeds up with angular acceleration...

Question

A Ferris wheel of radius R speeds up with angular acceleration starting from rest. Find expressions for the (a) velocity and (b) centripetal acceleration of a rider after the Ferris wheel has rotated through angle ∆θ.

Accepted Answer

Hi, everyone. In this practice problem, we're being asked to find the expressions of the spheres linear velocity and radial acceleration in terms of the alpha and the two minus the one, we have a sphere attached to a rod that is rotated starting from rest in a circular path of diameter D with a constant angular angular acceleration. Alpha. The sphere actually undergoes an angular displacement of theta two minus theta one. And we're being asked to find the expression for the spheres linear velocity and also radial acceleration. The options given are A V equals D multiplied by a square root of alpha multiplied by theta two minus theta one. A equals to D multiplied by alpha multiplied by theta two minus theta one B V equals to D multiplied by square root of open parenthesis. Alpha multiplied by theta two minus theta, one of that divided by two close parenthesis. And A equals to D multiplied by alpha multiplied by theta two minus theta one CV equals to two D multiplied by a square root of open parenthesis. Alpha multiplied by theta two minus theta one close parenthesis and A equals to D divided by two multiplied by alpha multiplied by theta two minus the one DB equals to square root of open parenthesis. D multiplied by alpha multiplied by theta two minus theta one, all of the divided by two close parenthesis and A equals to D divided by two multiplied by alpha multiplied by theta two minus theta one. So in order for us to actually find the fierce linear velocity and radial acceleration, what we wanna do first is to find the expression for the angular velocity. So in order for us to find that what we wanna do first is to look at the equation for the angular acceleration. So the rate of change of angular velocity of a particle rotating in a circular orbit is called the angular acceleration, which is represented by alpha. So alpha actually equals to delta omega divided by delta T where the omega is going to be the angular velocity. We wanna rearrange this in order for us to get alpha multiplied by delta T equals to delta omega. And we want to actually integrate both sides of this equation between the initial and the final condition. So we know that the initial condition is going to be at T equals to zero seconds and at T equals to zero seconds, the omega will equals to Omega knot which is going to be zero radiance per second. Which is it was because in the problem statement, it is mentioned that our sphere will be rotating starting from rest and at T equals T seconds. Omega T is going to equals to just omega radiance per seconds just like. So, so next, we want to integrate both sides. So integral of uh the omega starting from Omega knot to Omega T which is equals to omega and omega knot equals to zero radiance per seconds. Omega radiance per seconds. So integral of the omega starting from zero to omega radiance per second will equals to the integral from zero to T seconds of alpha D T. The right side will then equals two because the alpha is going to stay constant or the angular acceleration will always stay constant throughout zero till T we wanna pull that out of the integral. So alpha multiplied by the integral from zero to T of D T. That will essentially give us the equation for the left side. The left side is going to be omega minus zero equals to alpha multiplied by T. So that will give us Omega equals to alpha T. I am going to represent this with our first equation just so that it's easier for us to refer back to in the following calculation. So the angular philosophy omega of a particle is going to be the rate of change of the particle data as it moves in a circular trajectory or essentially omega well equals to delta theta divided by delta T. And we want to rearrange this equation so that we get omega delta T equals to delta theta. And we want to do exactly the same thing here just like as what we did previously with our first equation of angular acceleration. Initially at T equals to zero seconds, theta is going to equals to theta one while at T equals to T seconds, data is going to equal to theta two. So integrating both sides of the equation, we have the integral from theta one to theta two D T or D theta will equals to the integral from zero to T omega delta T. And we want to next substitute equation one into this equation right here in order for us to actually get Equals to the integral from 0 to T of Alpha T multiplied by PT. So the alpha will remains constant. So we can pull it out of, out of the integral. So this side will be equals to alpha multiplied by the integral from zero to T of T multiplied by D T. So we wanna recall that uh in the girl of X D X will equals to half X squared plus C which we are going to neglect the plus C in this case. And we can use this following integral rule in order for us to solve our equation. So the last side, the integral from theta two to theta one of the theta is then going to just be theta two minus theta one and that all equals to alpha multiplied by half of T squared, I'm going to rearrange this so that theta two minus theta one will equals to half alpha T squared just like. So And I'm going to call this equation two. So what we wanna do is to combine equation two and equation 11 in order for us to actually get omega without the time variable. So the way we want to do that is to by first rearranging equation two, in order for us to get an equation for time equals to something so that we can then substitute that equation for time into equation one. So based on equation two, we have theta two minus theta one equals to half alpha T squared. We wanna rearrange this so that we get T equals to the square root of two multiplied by theta two minus theta one divided by alpha just like. So, so now that we have the equation for time, we want to substitute our time equation into equation one that we have previously. So equation one here is Omega equals to alpha T. So that will give us omega equals to alpha multiplied by the square root of two multiplied by theta two minus theta one divided by alpha just like. So next, we want to simplify this and pull the alpha into our square root in order for us to then get the square root of two alpha multiplied by theta two minus theta one. So the final equation that we get is the equation for the angular velocity omega without time, without the, without the time variable, which is going to be equals to the square root of open parenthesis, two multiplied by alpha multiplied by theta two minus theta one close parenthesis. So next, we want to relate this to the relationship between linear and angular speed which is going to be V equals to R multiplied by omega. So R is going to be the radius of the circle which is going to be D divided by two. So essentially V will equals to D over two multiplied by omega. So the serious linear velocity will then equals two V equals to D over two multiplied by the square root of our omega, which is two alpha multiplied by theta two minus theta one. We can actually simplify this further by pulling the half into our square root. So that our linear velocity will then equals to D multiplied by the square root of alpha multiplied by theta two minus theta one, all of that divided by two still inside of the square root. So now that we have found our spheres linear velocity, what we can find next or what we have left to do is to find the Swedish radial acceleration. So we want to recall that the relationship between the spheres radial acceleration and the spheres angular speed is represented by A equals to R Omega square. The R is going to be D over two, just like how we calculate our linear velocity. So A will equals to D over two multiplied by omega squared. Now, we can substitute our equation for omega into our equation for our radial acceleration in order to, for us to get the expression for A. So A is going to equal to D over two multiplied by omega squared, which is going to be the square root of two alpha multiplied by theta two minus theta, one of that is going to be squared. So that will essentially cancel the square root and also the square together. So our A equation will then come out to just be D alpha multiplied by theta two minus theta one just like. So, so that will be the final answer or the final expression for the um radial acceleration, which is going to be D multiplied by alpha multiplied by theta two minus theta one. And the final uh expression for linear velocity, which is going to be D multiplied by the square root of open parenthesis alpha multiplied by theta two minus the one close parenthesis, all of that divided by two still inside of the square root. So that will essentially correspond to our option B in the answer choices. So option B is going to be the answer to this particular practice problem. So that will be all for this particular practice problem. 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.

A Ferris wheel of radius R speeds up with angular acceleration starting from rest. Find expressions for the (a) velocity and (b) centripetal acceleration of a rider after the Ferris wheel has rotated through angle ∆θ.

Key Concepts

Angular Acceleration

Tangential Velocity

Centripetal Acceleration

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