Question
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IP After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s (Figure 1). To reach the rack, the ball rolls up a ramp that rises through a vertical distance of 0.53 m. What is the linear speed of the ball when it reaches the top of the ramp? Express your answer using two significant figures. If the radius of the ball were increased, would the speed found in part A increase, decrease, or stay the same? increase decrease stay the same
Explanation / Answer
Given: The vertical distance, h = 0.53 m The linear speed of the ball, v = 2.85 m/s ------------------------------------------------------------------------------
Solution: Here as the ball is rolling, the rotational kinetic energy of the ball is taken into account The rotational kinetic energy of the ball is Krot = 0.5I?2 Where I is the moment of inertia of the ball = (0.4)MR2 Here, M is the mass of the ball R is the radius of the ball Therefore, the rotational kinetic energy of the ball is Krot = (0.5)(0.4MR2)(v/R)2 Here, v is the speed of the ball So, the above equation becomes Here, v is the speed of the ball So, the above equation becomes Krot = (0.2)Mv2 ---------------------------------------------------------------------------------
At the initial position: The potential energy, Ui = 0 The kinetic energy, Ki = (0.5)Mv2 The rotational kinetic energy, Krot,i = (0.2)Mv2
At the final position: The potential energy, Uf = Mgh The kinetic energy, Kf = (0.5)Mv'2 The rotational kinetic energy, Krot,f = 0.2Mv'2 ----------------------------------------------------------------------------------- Applying the conservation of energy 0.5Mv2 + 0.2Mv2 = Mgh + 0.5Mv'2 + 0.2Mv'2 0.7
v2 = gh + 0.7 v'2 Substitute the given data in above equation, we get (0.7)(2.85 m/s)2 = (9.8 m/s2)(0.53 m) + (0.7)v'2 Therefore, the linear speed of the ball when it reaches the top of the ramp is
v' = 0.838 m/s = 0.84 m/s (nearly) At the final position: The potential energy, Uf = Mgh The kinetic energy, Kf = (0.5)Mv'2 The rotational kinetic energy, Krot,f = 0.2Mv'2 ----------------------------------------------------------------------------------- Applying the conservation of energy 0.5Mv2 + 0.2Mv2 = Mgh + 0.5Mv'2 + 0.2Mv'2 0.7
v2 = gh + 0.7 v'2 Substitute the given data in above equation, we get (0.7)(2.85 m/s)2 = (9.8 m/s2)(0.53 m) + (0.7)v'2 Therefore, the linear speed of the ball when it reaches the top of the ramp is
v' = 0.838 m/s = 0.84 m/s (nearly) Applying the conservation of energy 0.5Mv2 + 0.2Mv2 = Mgh + 0.5Mv'2 + 0.2Mv'2 0.7
v2 = gh + 0.7 v'2 Substitute the given data in above equation, we get (0.7)(2.85 m/s)2 = (9.8 m/s2)(0.53 m) + (0.7)v'2 0.7
v2 = gh + 0.7 v'2 Substitute the given data in above equation, we get (0.7)(2.85 m/s)2 = (9.8 m/s2)(0.53 m) + (0.7)v'2 Therefore, the linear speed of the ball when it reaches the top of the ramp is
v' = 0.838 m/s = 0.84 m/s (nearly)