You are participating in league bowling with your friends. Time after time, you

Question

You are participating in league bowling with your friends. Time after time, you notice that your bowling ball rolls to you without slipping on the flat section of track. When the ball encounters the slope that brings it up to the ball return, it is moving at 3.41 m/s. The length of the sloped part of the track is 3 m. The radius of the bowling ball is 11.5 cm. (a) What is the angular speed of the ball before it encounters the slope? ? rad/s (b) If the speed with which the ball emerges at the top of the incline is 0.40 m/s, what is the angle (assume constant) that the sloped section of the track makes with the horizontal? ?

Explanation / Answer

The Kinetic Energy formula for linear motion is of course (1/2)mv^2 which is= to .5m(3.41 m/s)^2 The KE formula for angular is (1/2)(I)(w^2) I in this problem =(2/5) m r^2 so now our KE for angular in the beginning = (1/2)(2/5)(m)(r^2)(w^2) and to simplify further we know that w=v/r and so (w^2)=(v^2)/(r^2) and therefore find that our KE angular =.2mv^2 Add those two together and you have the beginning energy =.7m(3.41 m/s)^2 then we set that to the Final Energy The Final Energy is a combination of Potential Energy with h=height (which is what we want to find) and Linear Kinetic Energy (with a final v=.4 m/s) and Angular Kinetic Energy (since there is a final v larger than zero than there has to be a final w, and therefore an Angular Kinetic Energy component in the Final Energy) Final PE (Potential Energy) =mgh=(9.81 m/s^2)mh Final KE (Kinetic Energy Linear) =.5m(.4 m/s)^2 Final Angual KE =.5Iw^2=.5(2/5)mr^2w^2=.2mv^2=.2m(.4 m/s)^2 add them all up to get Final E = m(9.81h + .08 + .032)=m(9.81h + .112) set this Final E = Initial E to find h m(9.81h + .112)=m(12.28927) cancel out the m and solve to find h=1.24131 meters then use inverse sine to find that theta=27 degrees yay! hopefully i m not an idiot and that is the correct answer

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