The 'Giant Swing' at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a cable 5.00 m long, and the upper end of the cable is fastened to the arm at a point 3.00 m from the central shaft (Fig. E5.50). (a) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of 30.0° with the vertical.

Video transcript

Hey everyone today, we're dealing with the problem regarding centripetal acceleration. So we're being told that a merry go round at amusement park is made up of a vertical shaft with a number of horizontal arms attached to it at the top, the top end of a seven m long cable, seven m long cable is attached to an arm that is four m from the central shaft. And this arm suspends a seat from the lower end. If the cable holding a seat makes a 60° angle at the vertical, we're being asked to find the duration of one swing, one swing revolution. So there's a few formulas we need to know here. The first is the formula for centripetal acceleration, which states that the acceleration, centripetal acceleration is equal to V squared over R. Where he scored as the velocity and r. Is the radius directed horizontally to the left. Let's draw that in here too. This is centripetal acceleration. The other formula that will need and I will come back to is the period the time for one revolution or one swing revolution. A period is defined as two pi R over V. Where R. Is the radius and V. Is the velocity yet again. So first let's find the radius of the circle. We know that the arm starts from four m from the center and we know that the arm itself, the cable is seven m long. So we know that the angle between the ah Seat, the cable holding the seat and the vertical is 60°. And we know that the length of the cable is seven m. So that means we can make a sort of right triangle here. Right? And if we do so we can also find the remaining distance to the end the radius. So it'll be four m the radius. Right? This in blue, You know will stand black. The radius will be four m plus We have the seven m. But if we recall our trigonometry rules, if this is a right triangle and let me use our laser pointer for this. If we have a right triangle here then this leg will be congruent to this leg right here will be the same. Which means if we apply our So kato rules are trigonometry rules. We know that sign is the adjacent over the or sorry Sign is the opposite over the hypotenuse which means this length will also be signed 60 multiplied by the length of the hypotenuse. Once we simplify so this will be 7m times sine of 60° Giving a value of 10 m. With that in hand we can go ahead and apply Newton's second law and projecting along the vertical and horizontal axis gives us a couple of their forces as well. We have the force this way as well as the force due to gravity. And recall yet again with our trigonometry rules we have a force equal to MG. Right. But with our trigonometry rules again, the mass or the weight will be congruent to the force downwards here. This vertical component of the force which is opposite the hypotenuse or sorry adjacent to the hypotenuse. So we use co sign. So this will simply be F cool sign 60. And this also means that the F Science 60 will therefore be An F sign 60 will be M V squared over R. So what we need to do now is divide F Science um this term here by this term here and what this gives us if we divide them we get that sign theater over Kassian data is equal to V squared over G R. Where G is the um where G is the force of acceleration due to gravity. Now solving for V squared we get G R 10 Theta. Alright, this a little better. 10 theta, solving for V. We get V is equal to the square root of G R 10 theater which when we substitute in our values we get the square root of 9.8 m/s squared into 10 m into 10 60 Which gives us a final answer of 13.03 m/s. But we're not done. We still need to find a period. So the time for one revolution one revolution will be two pi r over v which is two pi into 10 m over 13.03 m/s giving a final final answer a 4.82 seconds. So the period. The time for one swing one swing revolution is 4.82 seconds. Or answer choice D. I hope this helps, and I look forward to seeing you all in the next one.