Suppose you swing a ball of mass m in a vertical circle on a string of length L. As you probably know from experience, there is a minimum angular velocity ωₘᵢₙ you must maintain if you want the ball to complete the full circle without the string going slack at the top. a. Find an expression for ωₘᵢₙ.

Video transcript

Hi, everyone. In this practice problem, we are asked to determine the minimum angular acceleration of rotation in order for our sphere to still complete a full rotation without the cable becoming loose at its highest position. So we were given a tiny sphere with a mass of 125 g revolving in a vertical plane around a horizontal axis at the end of a cable of length centimeter. So we are being asked to find the minimum angular acceleration of rotation so that the sphere can still complete a full uh a full rotation without the cable actually becoming loose at the highest position. The options given are a 2.8 radiant per seconds. B 3.5 radiant per seconds, C 8.8 gradients per seconds and D 12. radiant per seconds. So the sphere will undergo a vertical circular motion. And we're being asked to look at the position where the sphere is at its highest point. So at the highest position of its motion, the sphere will be subjected to two different forces. So this is going to be shown in our diagram right here. So in our diagram, the circular or, or the circle is going to represent the trajectory at its, at which the sphere is rotating at. And this is going to represent the cable holding the sphere. Now, we will also have the sphere itself at its highest point or at its highest position. And at this highest position, the sphere will have will be subjected to two different forces. The first one is represented by this red color which is going to be the weight, it is m multiplied by the G. And the second one is going to be the tension of the cable itself, which is represented by this blue color, which is going to be represented by our. At the same time, we know that the sphere is still moving to complete its full rotation or full revolution. And because of that, the sphere will have a velocity going perpendicular to the cable. Um assuming that the sphere is rotating counterclockwise, it is going to be going to the left or the velocity is going to be going to the left. So we know that the sphere is subjected to two different forces. The first one is the weight. The second one is the tension and both are um in the vertically downward direction. So at this position, we wanna apply Newton's second law in order for us to actually find what we're looking for, which is the minimum angular acceleration of rotation. So at the highest point along the radial direction. Using Newton's second law, we will get Sigma F R equals M multiplied by A R. The Sigma F R is the summation of the forces acting on that position. And the mass is the mass of our system and the acceleration is our acceleration in the radical or radial direction. Uh The forces acting on are going to be our weight plus the tension which will equals to M multiplied by A R. And we want to recall that R A R or radial acceleration is equals to P squared divided by R. And we wanna substitute that equation into our um Newton second law in order for us to actually get an equation for the velocity. So M G plus P will then equals to M multiplied by V squared divided by R. And this will be the equation that we are working with. So the next thing that we wanna be um taken in account of is because the tension D in the cable will not be able to be negative. In order for our tiny sphere to keep maintain and complete a full rotation or a full revolution. We thought the cable becoming loose at its highest position, then the tension must be then equals to zero Newton. This is uh again, considering that the tension in the cable will not be able to be negative. So at the tension of zero Newton, this will give us the minimum velocity or the smallest speed at which the sphere can revolve in a vertical circular motion without the string becoming loose. So if the speed is less than mean or the minimum Felos, if the speed is less than the minimum velocity, then the the string will either become loose or the sphere will not be able to complete its vertical circular motion. So in order for us to find the angular acceler acceleration of rotation off the sphere, when it's still, uh when it's still able to maintain a complete full rotation without the cable becoming loose, we need to find the minimum velocity. I'm gonna say the minimum velocity is uh essential for the sphere to maintain full rotation plus cable not Becoming loose. All right. So next, we can then uh substitute the tension value of zero Newton into our equation. So that our equation will then be M G equals to M multiplied by V squared divided by R. We have both MS on both sides so that we can, we can neglect that or uh cross that out. So our equation will then becomes G equals V squared divided by R. Next through rearrangement. We are interested in finding our philosophy so that we can then calculate our minimum angular acceleration. So V will then equals to the square root of G multiplied by R. And we want to recall that when RT equals to zero Newton, that is going to be our minimum velocity. So I'm just gonna add the VM or the min parted into our expressions. All right. So the relationship between the linear speed of velocity and the angular rotation is actually going to be given by the following expression. So v will equals two are multiplied by Omega rearranging. This Omega is going to equals to V divided by R. And alongside this Omega min is going to equals to V M divided by R with Omega being the angular rotation, which is the one that is being asked. So substituting the V min uh equation that we have, then the Omega min will then equals to the square root of G multiplied by R divided by R which uh is going to equals to the square root of G divided by R. So we can then substitute all the values that we know which is uh the G and the R which is the length of the cable in this case. So the G is going to be just the normal gravitational acceleration which is 9. m per second squared. And then the R is going to be 80 centimeters or 0. m. And that will give us uh omega minimum value of 3.5 radians per seconds. So the minimum angular acceleration of rotation is then going to be 3.5 radiance per second. And that will correspond to option B. So option B is going to be the answer to this particular problem and they'll be all for this particular example or practice. If you still have any sort of confusion, please make sure to check out our other lesson videos on similar topic and they'll be all for this one. Thank you.