a. An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ. Find an expression for the angular velocity ω.

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Hey, everyone in this problem, a sphere of mass M is attached to a rope of length, L divided by two, the sphere rotates in a circle at a constant speed. The rope is slanted downward at an angle of theta with the horizontal. And we're asked to derive an expression or the angular velocity omega of the sphere. In terms of L and theta, we have four answer choices here options A through D, each of them containing a different expression with GL and theta. And we're gonna come back to those once we're done working through this problem. And we're also given a diagram of the situation. So we have our sphere mass M attached to the rope of length L divided by two. And I'm just making an angle of the, with the horizontal. We can see that it's rotating in that circle. OK. So let's go ahead and draw out a free body die. If we draw a free body diagram of our sphere, we know that we have the force of gravity acting downwards. We have the tension acting up into the right now, our centripetal acceleration is going to be center seeking. So it is going to be pointing straight to the right in the center of the circle that we are making. Now, if we have our centripetal acceleration pointing to the right like this, then what we wanna do is go ahead and break up that tension into this interpret direction and then the upwards direction. OK. So we're gonna break that tension up into the TY and TX components. Now, let's think about this centripetal direction. I recall that the centripetal acceleration is related to the velocity and it can be related to the angular velocity. So if we use our forces to look at the centripetal acceleration, we are gonna be able to get omega the angular velocity. So our centripetal forces may recall that the sum of those forces is gonna be equal to the mass multiplied by the acceleration, the centripetal acceleration. Now, in our centripetal direction, the only force we have acting is TX, the X component of the tension. So we get that TX is gonna be equal to the mass. Now, the centripetal acceleration can be written as V squared divided by RK or V. Is that linear or tangential velocity? We don't want the linear velocity, we want the angular velocity. So let's recall that we can convert that as well. So the tension TX is gonna be equal to the mass multiplied by V which we can write as R omega all squared divided by a simplifying this. We know that the X component of tension, TX is equal to M multiplied by R multiplied by Omega squid. We wanna go ahead and solve for Omega. So we can divide by M and R. We get that Omega squared is TX divided by M. Now the X component of the tension, let me go to our diagram and the X component of the tension is going to be related to cosine or sorry related through cosine because the adjacent side is relating to that angle through cosine. So what that tells us is that Omega it's going to be equal to the square root ft cosine the divided by Mr All right, we're getting really, really close. When we look at this, we have an expression for Omega like we want it, we have T in our expression theta M and R right now, we don't want T in our expression. We want it in terms of L and theta. So we need to figure out what T is, how can we do that? Let's go back to our free body diagram and let's now look in the Y direction and what's going on in the Y direction. Well, the sum of the forces in the Y direction is going to be equal to zero. OK. Because this ball is not, or this sphere is not moving up and down, OK. It's moving around in a circle but it is not moving up or down vertically. And that means it's an equilibrium when we talk about the vertical direction, the sum of the forces is zero. Now, what are the forces in the Y direction? Well, let's take upwards to be our positive direction. We get ty acting in a positive direction FG acting in the negative direction. So ty minus FG is equal to zero, the Y component of tension is going to be related to the tension through sine of the angle. OK? Because now we're talking about the opposite side. So we get that T sin of the, it's going to be equal to the force of gravity which is the mass M multiplied by the acceleration due to gravity. G. So we can write the 10 T as MG divided by sine of data. So now we've been able to write the attention in terms of some of these variables that we want in our expression. Let's go back to our equation for Omega and substitute what we just spent. All right. So now Omega is going to be equal to the square of the 10 T which we just found to be MG divided by sine of the multiplied by cosine of the divided by M getting there. Let's simplify a little bit more Omega is going to be equal to, well, the mass M will divide them and we're gonna be left with the square root of GEO data divided by our signed data. Once again, we're getting closer, but we do not want this value of R in our equation, right? We wanna use L and we wanna use theta. So let's go up, let's look at our diagram again, we can see that L divided by two, that string and R form this triangle and they form two sides of the triangle with angle theta. So we can use some triangle math here to figure out R in terms of L and substitute that into our equation. Now, we're gonna do that on the right hand side. So if we draw this little triangle in and we have the angle theta, the hypotenuse is L divided by two and the adjacent side is R and we wanna relate the hypotenuse in the adjacent sides. We're gonna use cosine cosine of the is going to be equal to R divided by L divided by two, which tells us that our is going to be equal to L divided by two multiplied by cosign it. All right. Again, back to our equation for Omega. We're working through this bit by bit, figuring out what all of these variables are in terms of our variables of interest. When we go back to our equation for Omega, we now have that Omega is going to be equal to the square root og cosign of data divided by R which is L divided by two cosine of beta multiplied by sign of death. Those cosines are gonna divide it. We're gonna be left with Omega being equal to the square root of two G divided by L A theta. And that's exactly what we wanted. We wanted this expression for Omega in terms of GL and beta. And that's exactly what we have. If we go up to our answer choices, let's go ahead and compare each one of them. Now, option A is two G divided by L sine theta all to the exponent one half. I recall that the exponent one half is equivalent to the square root. And so option A actually matches what we found. OK. So option A is the correct answer, but let's go through the rest and just be sure that we've got it correct. Option B has G divided by two divided by two LS sign data to the exponent one half. OK? So those that coefficient in front of the numerator is not correct. Option C has the same function inside of the exponent but it's raised to the power of one third. OK. So that's not the square root, that's the Q root which is not correct. And option D has two things wrong. The coefficient in front of that numerator is missing and we have the incorrect exponent. We have a quarter here instead of A OK. So option A two G divided by L sine theta all to the exponent one half is the correct answer. Thanks everyone for watching. I hope this video helped see you in the next one.

a. An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ. Find an expression for the angular velocity ω.

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Key Concepts

Centripetal Force

Angular Velocity

Trigonometric Relationships in Circular Motion