Pts note the following three cases in which case will

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10. (6 pts.) Note the following three cases. In which case will the magnitude of the normal force on the block be equal to ( Mg + F sin θ )? A. Case 1 only B. Case 2 only C. Both cases 1 and 2 D. Both cases 2 and 3 E. Cases 1, 2, and 3 11. (6 pts.) The figure gives the angular momentum L of a wheel versus time t. Which of the following is true for the magnitudes of the torques in the intervals shown? B C L t B C A τ τ τ τ > > = C B A D τ τ τ τ > > > A A A 12. (6 pts.): A bullet is aimed at a target on the wall a distance L away from the firing position. Because of gravity, the bullet strikes the wall a distance Δ y below the mark as suggested in the figure. Note: The drawing is not to scale. If the distance L were half as large, and the bullet had the same initial velocity, how would Δ y be affected? A B C L t D 2 v gr = v gr = shared via CourseHero.com D B C A τ τ τ τ > > = C B A D τ τ τ τ > > > This study resource was shared via CourseHero.com
Problem 1 (25 pts.): A bowling ball of mass M and radius R ( treated as a uniform solid sphere ) is thrown along a level floor with kinetic energy K 0 . It is thrown so that initially (at t = 0) it slides with a linear speed ν 0 but does not rotate. As the ball slides, kinetic friction causes it to begin spinning. Eventually the ball begins to roll without slipping along the floor, with linear speed ν and angular speed ω , as shown below. The coefficient of kinetic friction between the ball and the floor is μ k . a) ( 15 pts. ) At what time t > 0 does the bowling ball begin rolling smoothly along the floor? What is the linear speed ν at that time? Express your answers in terms of ν 0 , g, and μ k , as needed. I = 2 5 MR 2 (solid sphere) Ma = f = μ k Mg a = μ k g (COM deceleration) I α = fR = μ k MgR α R = 5 2 ( μ k g ) (acceleration of rotation) v = v 0 + at = v 0 μ k gt (linear motion of COM) ω R = α Rt = 5 2 ( μ k g ) t (Rotation motion) Rolling condition: v = ω R at t = t r v 0 + ( μ k g ) t r = 5 2 ( μ k g ) t r = 5 2 μ k gt r v 0 = 7 2 μ k gt r t r = 2 v 0 7 μ k g b) ( 10 pts. ) What is the change Δ K in the bowling ball’s kinetic energy during this entire process ( i.e ., from sliding motion to rolling without slipping)? Express your answer in terms of K 0 . K 0 = 1 2 Mv 0 2 (initial K.E) and K r = 1 2 Mv r 2 + 1 2 I ω r 2 (rolling K.E.) K r = 1 2 Mv r 2 + 1 2 I ω r 2 = 1 2 Mv r 2 (1 + I MR 2 ) = 1 2 Mv r 2 ( 7 5 ) But v r = v 0 + at r = v 0 μ k gt r = v 0 ( 2 7 ) v 0 = 5 7 v 0 [from the result in (a)] K r = 1 2 Mv r 2 ( 7 5 ) = 1 2 M 5 7 v 0 ! " # $ % & 2 ( 7 5 ) = 5 7 K 0 Δ K = K r K 0 = 2 7 K 0 v o ω 0 = 0 f x r This study resource was shared via CourseHero.com

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