2.0 kg ball swings in a vertical circle on the end of an 80-cm-long string. The tension in the string is 20 N when its angle from the highest point on the circle is θ = 30°. a. What is the ball's speed when θ = 30°?

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Hey, everyone in this problem, a child is playing with the yo yo which has a mass of 0.5 kg and is attached to a spring that is 60 centimeters long. The child set the yo yo into motion by swinging it in a vertical circle. When the yo yo is at an angle of 45 degrees from the highest point on the circle, the tension is in the string is measured to be 25 new. We're asked to find the speed of the yo yo at this instant assuming that there is negligible air resistance and no loss of energy due to f we have four answer choices here. Options A through D all in meters per second. Option A 8.12. Option B 5.84 option C 3.29 and option D 4.61. We're gonna start by drawing a little picture of what we have going on here. OK. So we have this child playing with the yo yo, OK? You're gonna be holding the yo yo at a point and that yo, yo at the moment we're interested in is going to make a 45 degree angle with the highest point. OK. So with the vertical up, so that is a 45 sugary angle. They are told that this, yo, yo has a string that is 60 centimeters long. OK. So the distance between the point where the yo yo is rotating about and the end of the yo yo is 60 centimeters. And we're told also that the yo yo has a mass of 0.5 kg. OK? So there's a little picture of what's going on. And then this yo yo is getting spun around in a vertical c as you can imagine it going around and around and around. Now that we have this picture, we wanna go ahead and draw a free body diagram and think about the forces act. They were told about the tension in the string. So let's go ahead and draw that free body diagram and figure out what forces we have acting. So we have our y Yeah, and we're gonna have the force due to gravity acting on that yo yo straight downwards. And then we're gonna have the 10 force acting along that string on that 45 degree angle with the verp. Yes, we have the tension force t now our centripetal acceleration, can I recall? That's gonna be center seeking? So that's gonna point to the center in the same direction as tension. All right, one more thing we're gonna do here, the force of gravity and this tension force are not in the same direction, but they're not perpendicular. OK. So we're gonna break this force of gravity up into components and there's gonna be a force of gravity component that points in the exact same direction as 10. And we're gonna call that FGC. OK. So that's the force of gravity pointing in that centripetal direction. Now, let's think about our summer forces and we know that the summer forces, centripetal axis is gonna be equal to the mass multiplied by the centripetal acceleration. Now, you might be thinking, how is this gonna help us? We asked for the speed. Well, recall that the acceleration, the centripetal acceleration can be written as V squared divided by R and so on the right hand side of our equation, we have MV squared divided by R. Now we have that V value that we're looking for and we want to find the speed and that's exactly what V is. So now we have that speed in our equation. OK. Let's worry about the left hand side. Now, some of the forces in that centripetal axis, we have the tension force acting in the positive direction as well as that centripetal component of the force of gravity. So T plus FGC is equal to MV squared divided by R. Let's go ahead and isolate for V. So we're gonna multiply both sides by R divide by M. We get that V squared is equal to R divided by M Oops, let's give ourselves some more space there. B squared is equal to R divided by M multiplied by T plus F GC. Sorry. And to isolate completely. For V, we take the square root, we get that B is equal to the square root of R divided by M multiplied by T plus FGC. All right. Now we're ready to plug in our values. Now, the force of gravity, what's the force of gravity gonna be? Well, we know that the force of gravity is the mass multiplied by the acceleration due to gravity. G. In this case, we want the component in the centric direction. If we go back and look at our diagram, we can see that the makes a 45 degree angle. OK. Again, we were told that in the problem and it's gonna be related through cosine of the angle because it is the adjacent side, right? And so we have MG cosine of 45 degrees or that force of gravity in the centripetal. OK. Substituting in all our values V is going to be equal to the radius part. OK. Now this string is 60 centimeters long. So that radius is going to be 60 centimeters. Convert that to meters by dividing by 100 we get 0.6 m divided by the mass of 0.5 kg multiplied by the tension 25 newtons plus the mass, 0.5 kg multiplied by the acceleration due gravity 9.8 m per second squared multiplied by cosine of 45 degrees. And again, all of this is inside of that square. We work all of this out on our calculator. We get the speed V that's approximately equal to 5.844 4 7 m per second. Ok. And that is what we were looking for. We go back up to the answer choices and we can see that the speed of the yo yo at the instant that we were interested in is going to be option B 5.84 m per second. Thanks everyone for watching. I hope this video helped you in the next one.

2.0 kg ball swings in a vertical circle on the end of an 80-cm-long string. The tension in the string is 20 N when its angle from the highest point on the circle is θ = 30°. a. What is the ball's speed when θ = 30°?

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Key Concepts

Centripetal Force

Newton's Second Law

Kinematics in Circular Motion