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When an object moves a constant speed around a circle a force is necessary to keep changing its direction of motion. This is because any change in the direction of motion entails a change in the velocity of the object. This is because velocity is a vector quantity, and if the direction of a vector changes, then the vector and hence the velocity has changed. The acceleration of an object moving with constant speed v around a circle of radius r has magnitude v^2 / r, and the acceleration is directed toward the center of the circle. This results from a force directed toward the center of the circle. Such a force is called a centripetal (meaning toward the center) force, and the acceleration is called a centripetal acceleration.

If a 12 kg mass travels at three meters/second around a circle of radius five meters, what centripetal force is required?

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Your solution: v^2/r=3m/s/5m=1.8m/s^2…Fcent=12kg*1.8m/s^2. I did look at your solution for a result unit, the rest I figured out on my own.

Confidence rating: Pretty Sure.

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Given Solution:

The centripetal acceleration of the object is v^2 / r = (3 meters/second) ^ 2/(5 meters) = 1.8 meters/second ^ 2. The centripetal force, by Newton's Second Law, must therefore be Fcent = 12 kg * 1.8 meters/second ^ 2.

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Self-critique (if necessary): Is there not a central unit for Fcent…12kg*1.8m/s^2 seems like a “longhand” way of writing something else. I’m thinking something along the lines of Newtons.

That's it, but in order to use the units meaningfully in calculations you need to know the expression in fundamental units.

Self-critique rating:

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Question: `q002. How fast must a 50 g mass at the end of a string of length 70 cm be spun in a circular path in order to break the string, which has a breaking strength of 25 Newtons?

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Your solution: .05kg*v^2/.7m=25N v=sqrt.((25N*.7m)/.05kg) v=18.7m/s

Confidence rating: Pretty Sure.

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Given Solution:

The centripetal acceleration as speed v will be v^2 / r, where r = 70 cm = .7 meters. The centripetal force will therefore be m v^2 / r, where m is the 50 g = .05 kg mass. If F stands for the 25 Newton breaking force, then we have

m v^2 / r = F, which we solve for v to obtain

v = `sqrt(F * r / m). Substituting the given values we obtain

v = `sqrt( 25 N * .7 meters / (.05 kg) ) = `sqrt( 25 kg m/s^2 * .7 m / (.05 kg) ) = `sqrt(350 m^2 / s^2) = 18.7 m/s.

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Self-critique (if necessary): Well if the distance around a circle = 2*~3.14*.r, then 2*~3.14*.7m=4.4m….and if it goes 18.7m/s then it goes around 18.7m/s/4.4m or 4.25 times per second.

Self-critique rating: Pretty Sure

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Question: `q003. What is the maximum number of times per second the mass in the preceding problem can travel around its circular path before the string breaks?

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Your solution:

Confidence rating:

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Given Solution:

The maximum possible speed of the mass was found in the preceding problem to be 18.7 meters/second. The path of the mass is a circle of radius 70 cm = .7 meters. The distance traveled along this path in a single revolution is 2 `pi r = 2 `pi * .7 meters = 4.4 meters, approximately. At 18.7 meters/second, the mass will travel around the circle 18.7/4.4 = 4.25 times every second.

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Self-critique (if necessary):

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

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Self-critique rating:

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Question: `q004. Explain in terms of basic intuition why a force is required to keep a mass traveling any circular path.

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Your solution: I believe this can be explained by the laws of inertia, an object in motion will stay in motion unless acted on by an outside force. All the forces acting on a ball rolling in a circle, for example, balance out the pull of the middle and push towards going outside of the circle.

Confidence rating: Pretty Sure

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Given Solution:

We simply can't change the direction of motion of a massive object without giving it some sort of a push. Without such a force an object in motion will remain in motion along a straight line and with no change in speed.

If your car coasts in a circular path, friction between the tires and the road surface pushes the car toward the center of the circle, allowing it to maintain its circular path. If you try to go too fast, friction won't be strong enough to keep you in the circular path and you will skid out of the circle.

In order to maintain a circular orbit around the Earth, a satellite requires the force of gravity to keep pulling it toward the center of the circle. The satellite must travel at a speed v such that v^2 / r is equal to the acceleration provided by Earth's gravitational field.

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&#Your work looks good. See my notes. Let me know if you have any questions. &#