A 2.0 kg pendulum bob swings on a 2.0-m-long string. The bob's speed is 1.5 m/s when the string makes a 15° angle with vertical and the bob is moving toward the bottom of the arc. At this instant, what are the magnitudes of (c) the tension in the string?

Video transcript

Hey, everyone in this problem, we're asked to consider a scenario in which a square metallic object of mass 1 kg hangs from a 3 m long rope. If the object is the space from equilibrium and allowed to oscillate, moving towards the bottom, it reaches a speed of 3.2 m per second. It makes an angle of 25 degrees with the vertical. We're asked to determine the tension in the rope at this particular instant. So we're given our diagram here. We have this square metallic object hanging from our rope making this 25 degree angle with the Berkel. We have four answer choices all in Newton's option, a 3.26 option B 6.28 option C 12.3 and option D 8.87. So let's go ahead and draw a free body diagram. OK. We're looking for the tensions. We want to think about the forces here. So if we draw a free body diagram of our block here, we have our block, we're gonna have the force of gravity acting downwards on the block. We have our attention acting upwards on an angle at that 25 degree angle from the vertical. Now, the direction of the tension is gonna be the direction of our centripetal acceleration. OK. Remember centripetal acceleration center seeking. And so that's pointing towards the center towards the pivot. OK. Or that point where this object is rotating a bit. All right. So our force of gravity, we wanna break up into this direction as well and we want to find the 10 and there's gonna be some amount of this gravitational force that acts in that direction. So we're gonna decompose that force of gravity. OK. And we're gonna call this component that points in the opposite direction of tension FGC. So the force of gravity in this centripetal direction. Now let's look at our sum of forces, we're called the sum of forces in centripetal direction. It's gonna be equal to the mass multiplied by the centripetal acceleration. Our forces in the interpretable direction, we have the tension in the positive direction and that force of gravity, the centripetal component in the negative direction. So T minus FGC is gonna be equal to the mass M multiplied by the acceleration ac. OK. So we're looking for the tension, we know the mass, we don't know the acceleration and we need to kind of fill in what this gravitational force is gonna be. So let's go ahead rearrange sulfur T and expand some of these terms. We get that the tension T is going to be equal to the mass multiplied by this interpretable acceleration, we recall that we can write this in interpretable acceleration as V squared divided by R and doing that is gonna be really helpful because we know the radius and we know the speed we were given those the problem. OK. So we went from this in Trippet acceleration which we didn't know to two values that we do know. And we're also gonna add FGC to this equation. We're moving it to the right hand side. Now, the force of gravity is given by the mass multiplied by the acceleration due to gravity. G. And then we're gonna multiply by cosine of 25 to get the component in that centripetal direction. All right, let's go ahead and substitute in our values. We get that the tension is equal to 1 kg multiplied by 3.2 m per second squared, divided by 3 m plus 1 kg multiplied by 9.8 m per second squared multiplied by cosine of 25 degrees. And when we work all of this out, we get that the tension is going to be approximately 12.295 noons. And that's exactly what the question was asking. If we go back up and look at our answer choices. We're going round to one decimal place and we can see or sorry, not one decimal place but three significant digits. And we can see that the correct answer is going to be option C 12.3 mutants. Thanks everyone for watching. I hope this video helped see you in the next one.