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QUESTION 9ball is being swung faster and faster in horizontal circle of radius 15 cm: When the speed reaches 20 m/s the tension (T) in the string is 500 N and the s...

Question

QUESTION 9ball is being swung faster and faster in horizontal circle of radius 15 cm: When the speed reaches 20 m/s the tension (T) in the string is 500 N and the string breaks; What is the mass of the ball?A. 3.75 kgB. 18.75 kg187.5 gD.375.0 kgQUESTION 10The passengers in car driving over hill at a speed of 12 m/s feel weightless Calculate the radius of curvature of the hillA. 144.0 mB. 12.0 mC.14.67 mD.1.22 m

QUESTION 9 ball is being swung faster and faster in horizontal circle of radius 15 cm: When the speed reaches 20 m/s the tension (T) in the string is 500 N and the string breaks; What is the mass of the ball? A. 3.75 kg B. 18.75 kg 187.5 g D.375.0 kg QUESTION 10 The passengers in car driving over hill at a speed of 12 m/s feel weightless Calculate the radius of curvature of the hill A. 144.0 m B. 12.0 m C.14.67 m D.1.22 m



Answers

An athlete swings a 5.00-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.800 m at an angular speed of 0.500 rev/s. What are (a) the tangential speed of the ball and (b) its centripetal acceleration? (c) If the maximum tension the rope can withstand before breaking is 100. N,
what is the maximum tangential speed the ball can have?

So let's draw our circle first. So this is the circle. This is the center off the circle, and we need to find the velocity off the man. Well, Felicity off the mast at the lowest point on the circle. So we're here and there is a string attached to Thomas on the limp off the string, which is in fact, the radius off the circle is given Toby One meter. I must off. The object is going to be one Katie and tension on the string. When the mass it more when the monster reaches the bottom is given Toby 100 den mutants No sins. The waas is moving in a circle. Therefore, there has to be a centripetal force on dhe the centripetal force. Let's call that FC is basically the some off the forces acting on the mass on dhe, it is always directed towards the center, off circle and motion. So if this is the moss, then the centripetal force at the bottom is goingto be towards the center. If the mass is over here, it's going towards the centre. Disintegrated before should be like this. So it's the force. Centripetal food's always points or is always directed towards the center, off the circle and motion. So if C is there, then gravity is downward, obviously, So that is mg on. The tension is upward. So let me just Oops, let me just it. Is this on right? Be over here because FC is basically a resultant force. Or is the net force on IFC comes into account as action off other external forces. So over here, F C comes into account because off the tension often on the street. Now let's drive down Newton's second law. So let me just pick another color. So Newton's second law says that neck forces a goto mass times acceleration. Now, over here, since the object is moving, Miss Circle the net force mass times acceleration. The acceleration is going to be the centripetal acceleration. Because the Net force is the centripetal force. The next words gives rise to centripetal force as the masses moving around in a circle on DDE. The centripetal acceleration is b squared over r. That artist radius off the circle and V is the velocity off the mosque at that point. Now let me light the net force acting on the mosque. So we have to be going the worst. The center of the circle there for this. That should be positive because I was centripetal. Force owns also goes towards the circle so that if the net force is towards the circle, we should If the net forces towards this our country, it means that the force external force that the stewards to circle should be dominant. I'll be over here. Then you have mg in the opposite direction, hands minus empty. And this is it will do last time centripetal acceleration and hymns and these guard over us. So from here you can solve for three squares, so that would be going toe do you minus mg names of over m. So in order to find the magnitude off V, we will take a square road off the whole expression on substituting all the values over here along with gravity. That stand in need of a second skirt will find the velocity at the lowest point off the circle. Toby 10 meet up a second. So the crack down service going, Toby Option B

Question Number seven asks us to find the tension in a string attached to a ball in two cases. In the first case, the bowl is at the top of its circular path, so let's draw the ball here. The string is below it and it's traveling at some velocity V as it is being swung around by this string. Now at the top of the ball's path. It's being acted upon by the force of tension in the rope that swinging it and the force of gravity. Some of these two forces will give us the centripetal acceleration of the ball. Also important is the length of our string, which is the radius R of the circle in which the ball travels are is measured to be 72 centimeters in length or 720.72 meters, and the velocity of the ball is given as four meters per second. Were also given the balls mass as 0.3 kilograms. Now, as I stated previously, the force of tension in the string Sebti plus the mass of the ball times the gravitational constant will be equal to the balls. Centripetal acceleration M times v squared over R. Now we want to find the force of tension in the string at this point. So let's subtract mg from both sides. Now if some tea will be equal toe MV squared over R minus MGI Now we can plug in all our Constance. Our force of tension will be equal 2.3 kilograms times four meters per second squared over 0.72 meters, minus 0.3 kilograms. Times are gravitational constant of 9.8 meters per second squared. Simplifying things down a little bit, We find that our force of tension will be equal to 6.67 nutrients minus 2.94 Newtons, which yields a value of 3.73 Newtons for strings tension when the ball is at the top of its path. The second part of our problem asks us to calculate the tension when the bull is at the bottom of its path. So now let's draw the ball at the bottom of its path with the string attached to it. It would have a velocity V going in this direction and the force of gravity and the force of the ropes. Tension would still be acting upon it, however, if they would be acting in opposite directions. Now, the force of tension would be pulling the ball upwards while the force of gravity hate would be acting in the opposite direction. Pulling the ball down, we can assume that the length of the string are is still the same as well as the velocity for that matter. However, now force attention minus M G will be equal to our centripetal acceleration. M times v squared over r Once again, Let's put em G on the other side so we get f some tea by itself. Now if some tea will be equal toe M fees squared over R plus m g. The previous part of the problem showed us that MV Squared over R was equal to 6.67 Newton's and MG was 2.94 Newtons, adding those two together, we find that the tension in the string when the ball is at the bottom of its path is equal to 9.61 Newtons

2 40 The mass off the bowl is Mullan. Virgin Pick was to find 100 grand. We have to calculate the gravitational force off the mark. Pollution full part A. We have the gravitational force on any object is equals. Toad of Louis equals toe energy. There g is 9.8 meters per second squared with the acceleration due to gravity. Now the mass off the ball is 500 grams now after converting it into a kilogram bigot 0.500 kilograms as the wheat is given by you don't want 500 kilograms. Multiply with 9.81 meter, four seconds. Look, we get 419 vehicle muted. The gravitational force on the while is 94.9 diluted. No bark be. When the bullets talk, the centrifugal force will act were tickly upward and the weight off the ball will act vertically. Don't work. There will be no force acting along horizontal direction. Hence the tension on the string will be difference between centrifugal fourth and wit. The weight off the ball is mold of Louisa Questo 4.90 Neutered Understand refusal force in given by F R, it was toe MV squared upon our Here are the radios off the part in case studies off the circular parties. It was to rent off Dustin, as we have our is questo l, which is it was 20.2 meter now substituting the value in this equation we get as our edict was too. Then difficult forces because to 0.500 kilograms, multiply with the willow 34 point ville meter fall second square divide by the ideas one want the row two meter. We have 7.84 Newton and their attention. Hence the tension on the string is given by teasing was to ask on my nest of you statistic was to 7.84 mutant minus 41 90 To neuter. We get 2.94 neutral as when the ball is that top off the 10 Trento. The tension of the string is toe 1 94 neutral Farbstein. The velocity of the ball at the bottom of the circle is redish, which is it was to 7.5 meters per second. We have to calculate the attentional destry at the bottom at the bottom the both waiting centrifugal force will act down work. Also, there is no horizontal force acting on the ball at the position as the tension of the string will be some off the reagents and difficult for third. This position does centrifugal force at this position if after fish are it was toe m vida squared, divided by R, which is substituting the value force and difficult force, we have 0.500 kilograms, four months. Willow 13. We have 7.5 return for second school meters per second bulls with divided by one plant the dough. To meet it, we get 27.6 noted. Hence, the attentional destry is the dishes it was too. After shot plus w between equals to 27.6 Newton, plus the blue 4.90 Newton we get pathetic was to 32.5 Newton. Hence the attention of the string at the bottom of the party's 32 +15 neutered

So here is the circular motions. Let's draw a circle now at the top of the circle, attention is acting downward. Also, the gravity acts downwards, so gravity here as well on dhe. Therefore the centripetal force which is the net force or some off all forces acting on the Mass and two worlds. The center is going to be some of these two forces of the centripetal force will be equal to the gravitational force. That's attention and center pretend forces M v squared over r So it's a student and gravitational forces just the weight off doc ticked less tension. No mas radius and gravity are all fixed valuables that cannot be changed. However, as the speed off the circling Mars is reduced, the tension in the spring of the tension the string will decrease. Now, at the minimum speed, the tension is going to be zero at the exact instant that is the mosque being at the top off the loop. So I back point attention is going to be able to zero because the minimum speed, of course, So we can say that in this guy overall is equal to most times gravity and therefore V which is the velocity off the speed at the top off the oh top of the circle of motion Top off the loop is going to be towed over off all times. T So we substitute the values the radius off the low pitches one neither gravity's. Then meet oppa Seconds. Girl, take a look over off that and find the minimum street that were that we were looking for is over off in need of a second. So the correct answer is option C Thank you.


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