A small mass m attached to the end of a string revolves in a circle on a
Question:
A small mass m attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table (Fig. 8-62). Initially, the mass revolves with a speed v1 = 2.4 m/s in a circle of radius r1 = 0.80 m. The string is then pulled slowly through the hole so that the radius is reduced to r2 = 0.48 m. What is the speed, v2, of the mass now?
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The area of a circle is given by the formula: A = ?r^2 Where A is the area, ? is a mathematical constant (approximately 3.14), and r is the radius of the circle. The radius is the distance from the center of the circle to the edge. So for example, if the radius of a circle is 5 units, the area would be: A = ?(5^2) = ?(25) = 78.5 square units Here three friends went out to a restaurant and decided to have pizza for lunch. While they were waiting for their order, one of them stepped out to take a call but promised to return in two minutes. The waiter came and presented them with the menu, and they decided to order two 10-inch pepperoni pizzas. The waiter took their order and left. Just then, the third friend returned and asked what they had ordered. They explained that they had ordered two small pepperoni pizzas as they were only three of them and it would be enough for them. However, the third friend suggested that for the same price, they could get a large pizza that would be much bigger than the two small ones. He asked the waiter for a pen and paper and proceeded to calculate the area of the two small pizzas and compared it to the area of one large pizza. The friends were amazed at how the circle measurements worked and decided to change their order from two small pizzas to one large pizza.
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