A particle of mass m is constrained to move on a circle of radius R. The circle
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A particle of mass m is constrained to move on a circle of radius R. The circle rotates in space about one point on the circle, which is fixed. The rotation takes place in the plane of the circle and with constant angular speed w. In the absence of a gravitational force, show that the particle’s motion about one end of a diameter passing through the pivot point and the center of the circle is the same as that of a plane pendulum in a uniform gravitational field. Explain why this is reasonable result.
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The xy coordinates of the particle are Then m 10 y x R cos at R cos cot y Rsin oot R sin co...View the full answer
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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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