This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: PHY 126-07 Mid-term Exam I Problem 1 (10+10 points) R (a) As the string exerts a force in form of tension on the puck in radial direction, there is no torque. Therefore the angular momentum is conserved (2 points) . The initial angular momentum L i =I i i (1 points) where I i and i are the moment of inertia of the puck mR 2 (2 points) and the initial angular velocity of the puck. The final angular momentum L f =I f f (1 point) where I f and f are the final moment of inertia m(R/3) 2 (2 points) and the final angular velocity, respectively. Therefore from L i =L f f / i =I i /I f =9. So, the angular velocity increases by a factor of 9 (2 points) . (b) The tension of the string T is given by Newtons 2 nd law, T=m a rad =mv 2 /r =(mvr) 2 /(mr 3 )=L 2 /(mr 3 ) where a rad is the radial acceleration of the puck, v the tangential velocity of the puck, r is the radius of the pucks orbit , and L=mvr is the angular momentum (5 points) . Since the angular momentum is conserved (2 points) , when r decreases by a factor of 3, the tension, therefore, increases by a factor of 27 (3 points) . Or a decrease by a factor of 1/27. Problem 2A (5+5+10 points)...
View Full Document