course Phy 201 ???{q????????E??Student Name:
.................................................
......!!!!!!!!...................................
19:47:58 `q001. Note that this assignment contains 4 questions. . Note that this assignment contains 4 questions. 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?
......!!!!!!!!...................................
RESPONSE --> The centripetal acceleration of the object will be v^2/r= (3m/s)^2 / 5m=1.8m/s^2. This, multiplied by the mass of the object, will give the centripetal force: 12kg*1.8m/s^2= 21.6N
................................................
......!!!!!!!!...................................
19:50:09 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:53:43 `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?
......!!!!!!!!...................................
RESPONSE --> The centripetal force would have to be 25N and would be =m*Acent=(.05kg)(v^2/.7m) V^2=(.7m*25N)/.05kg) v= 59.16m/s
.................................................
......!!!!!!!!...................................
19:55:18 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.
......!!!!!!!!...................................
RESPONSE --> When I did my calculation I accidentally used 7m for my radius instead of .7m. This gave me a velocity much too high which would be expected with such a long radius.
.................................................
......!!!!!!!!...................................
20:05:56 `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?
......!!!!!!!!...................................
RESPONSE --> I believe the answer you are looking for is circumference of the circular path=2pi*r=2(pi)(.7m)=4.4m. If the mass is traveling at 18.7 m/s, then in one second the mass will travel the circular path: 18.7 m/4.4m=4.25 times.
.................................................
......!!!!!!!!...................................
20:08:18 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.
......!!!!!!!!...................................
RESPONSE --> ok. This is what I did, however, I discussed this problem with my father and he agrees: if the string has a breaking force of 25N and therefore the string will break at 18.7m/s, it will not be traveling at this velocity. This is the velocity at which the string can no longer support the force of the mass. This problem implies that it is not a uniform velocity, but an increasing velocity approaching the breaking force. Therefore, the mass will not be traveling at this approached velocity.
.................................................
......!!!!!!!!...................................
20:09:46 `q004. Explain in terms of basic intuition why a force is required to keep a mass traveling any circular path.
......!!!!!!!!...................................
RESPONSE --> To travel a circular path, there must be something preventing the object from simply reacting to m omentum and continuing in a straight line. The force is what constantly changes the direction of the vector. The constant change of the y component of the vector will keep the mass in a circular path.
.................................................
......!!!!!!!!...................................
20:10:15 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
"