* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Solutions - faculty.ucmerced.edu
Survey
Document related concepts
Brownian motion wikipedia , lookup
Classical mechanics wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Atomic theory wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Equations of motion wikipedia , lookup
Specific impulse wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Mass in special relativity wikipedia , lookup
Classical central-force problem wikipedia , lookup
Work (physics) wikipedia , lookup
Electromagnetic mass wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Center of mass wikipedia , lookup
Transcript
Physics 8 Spring 2012 TA: NAME: Quiz 4 - Solutions Make sure your name is on your quiz, and please box your final answer. Because we will be giving partial credit, be sure to attempt all the problems, even if you don’t finish them! A mass m1 undergoes circular motion of radius R on a horizontal frictionless table, connected by a massless string through a hole in the table to a second mass m2 . If m2 is stationary, find expressions for 1. the string tension, 2. the velocity of m1 , and 3. the period of the circular motion (express your answer in terms of the masses, m1 and m2 , g, and the radius of the circle, R, and not the velocity, v of m1 ). ———————————————————————————————————— Solution 1. The tension in the string comes from the hanging mass. The spinning mass provides support against m2 falling any further down, but that’s simply the origin of the tension. Looking at the hanging mass, m2 , the only forces acting on it are the weight downward, −m2 g, and the tension, T , up. Balancing the two gives T = m2 g. 2. The weight of the hanging mass, m2 , provides the centripetal force that the spinning mass, m1 , needs to keep it from sliding off the table due to its inertia. So, the weight from the hanging mass is m2 g, while the centripetal force is m1 v 2 /R, where v is the velocity of m1 . Setting the two equal gives m1 v 2 = m2 g. R Thus, the velocity is r v= 1 m2 gR. m1 3. The period of the motion is T , and is simply the distance that m1 goes (the circumference, 2πR), divided by how fast it goes around (v found above). So, T = 2πR/v. Plugging in the values we found in question 2 above we find s r 2πR m1 m1 R = 2π . = 2πR T = v m2 gR m2 g 2