A 50 g mass is attached to a light, rigid, 75-cm-long rod. The other end of the rod is pivoted so that the mass can rotate in a vertical circle. What speed does the mass need at the bottom of the circle to barely make it over the top of the circle?

Question

Accepted Answer

Hey, everyone. So this problem is dealing with conservation of energy. Let's see what they're asking us. A physics student built a rotating device using a 1.1 m bar of negligible mass and a 75 g sphere fixed a fixed to one end of the bar. The opposite end is supported by a fixed fraction less horizontal axis, making it possible to freely rotate the bar in a vertical plane. Find the minimum speed that the student should give to the sphere at its lowest position so that the sphere makes a full rotation. Our multiple choice answers here are a 4.6 m per second. B 6.6 m per second, C 14.6 m per second or D 21.1 m per second. So the first thing I'm gonna do here is draw a little diagram kind of showing what's going on. So we have this bar, we have a sphere fixed to the end of it and it is going to rotate about that axis. And this bar we're told him the problem is 1.1 m. So from the conservation of energy equation, we know that the some of the kinetic and potential energy at one point is equal to the sum of the kinetic and potential energy at another point. As long as there are no additional outward forces acting on this system. So the student is going to push the sphere give it a speed at its lowest position. And then that sphere is gonna make a full rotation. So in using our conservation of energy equation that looks like the kinetic energy at the bottom of the rotation plus the potential energy at the bottom of the rotation is equal to the kinetic energy at the top of the rotation plus the potential energy at the top of the location, we can recall that our kinetic energy Is given as 1/2 MV two. So that will be 1/2 um V B squared. And then our potential energy is given as M G H. So M G H B the bottom and then at the top, we'll have the same one half M V top squared plus M G H at the top. So let's take, let's take a look at each of these terms. The first thing we can see here is that the mass is in each of these terms. So we can cancel that out. So the mass of the sphere doesn't actually matter here. That's interesting. We are solving four V B that's the speed at the bottom. OK. So G as our gravitational constant, we can recall that, that equals 9.8 m per second. Squared HB, the height at the bottom is going to be zero m. We'll assign that A height of zero m. Our speed at the top Is going to be zero m/s because we are looking for the minimum speed. So that means that at the top, it just barely makes it across that apex. So That speed is essentially zero and then HT is equal to two multiplied by the length of the bar And that equals 2.2 m. So as the bar Rotates around the bottom at the bottom, it is um the radius from the, from the fixed point is 1.1 m. And so when it makes its way to the top, it will be at 2.2 m away from where it was at the bottom. And from there, we can continue simplifying. So H B zero m. So this whole term actually goes to zero and VT, the speed at the top is zero so that the whole term goes to zero. So we are left with one half V B squared equals G H T. We have everything we need to solve for V B. So we'll rearrange and isolate that, that will be B B equals the square root of two G H T plug in those numbers two, multiplied by 9.8 m per second squared, Multiplied by 2.2 m. Plug that into our equation or into our calculator, excuse me. And we get 6.6 m/s. And so that is the answer to this problem. And that aligns with answer choice B so that's all we have for this one. We'll see you in the next video.

A 50 g mass is attached to a light, rigid, 75-cm-long rod. The other end of the rod is pivoted so that the mass can rotate in a vertical circle. What speed does the mass need at the bottom of the circle to barely make it over the top of the circle?

Verified Solution

Key Concepts

Centripetal Force

Gravitational Potential Energy

Conservation of Energy