A ball rolling on a circular track, starting from rest, has angul...

Question

A ball rolling on a circular track, starting from rest, has angular acceleration α. Find an expression, in terms of α, for the time at which the ball's acceleration vector a is 45° away from a radial line toward the center of the circle.

Accepted Answer

Hey, everyone in this problem, a sphere initially at rest is spun in a circle of an angular acceleration, alpha, the sphere's acceleration vector changes as a function of time. And we're asked to determine the time when the angle between the radius and the acceleration vector is beta degrees. And we're told to express T in terms of alpha and beta. And we have four answer choices for this problem. Option A the square root of alpha 10 theta. Option B is the reciprocal of option. A option C is one divided by alpha tan theta and option D is one divided by alpha tan theta squared. So let's start by kind of visualizing what's going on here. So we have a sphere that spun in a circle. OK. So we're gonna draw the center of our circle. OK. And we have a radius we're gonna draw as a dotted line out to our sphere. What's our radius? R we're interested in the angle between the radius. OK. And the acceleration vector. Now, when we're thinking about acceleration and we have something that's spinning around, we have to think about different types of acceleration. So we're gonna add two accelerations to our little diagram here. So we have the radial acceleration. OK. Sometimes this is in tal acceleration that it's called and that's gonna point from our sphere into the center of our circle along the radius. OK. So this is a radial acceleration which we call A R and then we have our tangential acceleration. I just gonna point upwards at the point that we've drawn. OK. It's gonna be the tangent to the sphere. So when the sphere is completely to the right, this tangential acceleration is pointing upwards, we're gonna call that A T. All right. Now, the spirit is spin and we have some angle between these two accelerations and that's the angle theta that we're told about. OK. The angle between the radius and the acceleration vector. OK. So we have our tangential acceleration and that radius vector or the radius lies along the radial acceleration. OK? They're the same line here. And so that angle theta is going to be between those two acceleration vectors. So if we consider this little triangle we've drawn, we can see that these two accelerations are gonna be related through the tangent. OK. So we can say that tangent of data which we know through our trig ratios is gonna be equal to the opposite side divided by the adjacent side. Well, that means that tan theta is gonna be equal to the tangential acceleration A T divided by the radial acceleration A R. And this tells us that we can write our tangential acceleration A T A as A R multiplied by 10 beta. Now, you might be looking at this and thinking, why are we doing this? How does this relate to the problem? OK. Now, remember we're trying to find the time when the angle between the radius and acceleration is theta degrees. So we've managed to work our theta degrees into this problem. OK? And now we need to relate it to the time T. So let's go to the side and work out some information about the accelerations we have in our approach. OK. So we have this tangential acceleration A T and we're called that that's gonna be equal to R multiplied by. OK. So with the radius multiplied by the angular acceleration. OK. So now writing that tangential acceleration, we're able to get this value of alpha into our equation. We're gonna do the same for the radial acceleration and the radial acceleration A R. Well recall it, this is gonna be V squared divided by R. Now we have some equation for A R we still need to introduce type. Yeah. So let's think about this. We know that the speed be, is gonna be equal to R omega. OK. So we have V in our A R equation. Now we can relate it to ome right. We can write our equation in terms of omega. And if we think about our kinematic equations or rotational motion, we can find what Omega is in terms of the other variant. OK. So let's write out the variables we have, we have Omega knot and we know that this is gonna be equal to zero if the sphere is initially at rest. Now for Omega F, which we're just gonna call omega, OK. The final angular speed. Well, this is what we're looking for. We have alpha and we don't know the numerical value of it, but we're told to use that variable in the problem. And delta T, well, this is what we're looking for. OK? We're just gonna call that T, we're gonna start at T equals zero. So delta T, we're just gonna call T. Now we have Omega not omega F and A, we're looking for the time T, OK. We have four equations for and not. So we can write our kinematic equation Omega equal Omega plus Albert Omega, not it zero. So we have that Omega is equal to alpha. Now, this is the key part. We've written an equation in terms of B, we've said that that can be written in terms of omega instead. And now we've written Omega in terms of the variable we're interested in T, OK. So this is how we're gonna relate our equation to that time T that we're interested in. So let me write A R now. OK? Now we can write our radio acceleration A R as B OK, which is gonna be R Omega. So R alpha, T all squared divided by R. OK. So let's go back to the equation we had, we had that the tangential acceleration A T was equal to the radial acceleration A R multiplied by 10. We now have an expression for A T. We have an expression for A R. We're gonna substitute those in, we have R alpha is equal to R alpha T squared divided by R multiplied by data. We can multiply both sides by R. We need to eliminate that denominator. We have R squared alpha gonna be equal to, we have R alpha T all squared. OK? So we can square each term and then multiply them together. So we get R squared alpha squared T squared multiplied by tangent of data. Now these are square terms and we can divide those out. Those are gonna cancel, we want to solve for teeth. OK? We wanna find the time. So we can say that T squared is gonna be equal to, well, we have alpha divided by alpha squared. So that's gonna give us one divided by alpha and then we're dividing by tangent of beta as well. So we have one divided by alpha and beta. OK? And we have to be careful here. On the left hand side, we have T squared, we want T, so we need to take the square root, we can go ahead and do that. And we get that T is gonna be equal to one divided by the square root of alpha 10 beta. So that is the time that we were looking for. That's the time when the angle between the radius and the acceleration vector is going to be this value of data. If we compare this to our answer choices, we can see that this corresponds with answer choice B Thanks everyone for watching. I hope this video helped see you in the next one.

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A ball rolling on a circular track, starting from rest, has angular acceleration α. Find an expression, in terms of α, for the time at which the ball's acceleration vector a is 45° away from a radial line toward the center of the circle.

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