Problem 10.60 - Enhanced with Feedback Part A You may want to review (m-pages 31

Question

Problem 10.60 - Enhanced with Feedback Part A You may want to review (m-pages 319-320) If the ball is released from rest at a height of 0.83 m above the bottom of the track on the no-slip side, what is its angular speed when it is on the frictionless side of the track? Assume the ball is a solid sphere of radius 2.7 cm and mass 0.14 kg.(Figure 1) Express your answer using two significant figures. 116 rad/s Submit My Answers Give Urp Incorrect Try Again; 3 attempts remaining; no points deducted You have used an incorrect moment of inertia in the expression for the rotational kinetic energy. Note that the ball is a solid shere, not a hollow sphere, so its moment of inertia has a numerical coefficient different from 2/3. Figure 1 of 1 Part B How high does the ball rise on the frictionless side? Express your answer using two significant figures. Submit My Answers Give Up No-slip

Explanation / Answer

Given

ball (solid sphere) of radius r = 2.7 cm , mass m = 0.14 kg

height h1 = 0.83 m

as the ball moving towards the frictionless surface via the lower point(horizontal) the gravitational potential energy converts in to rotationa k.e and translational k.e

that is mgh1 = 0.5*I*w^2 + 0.5*m.v^2

and rolling with out slipping is V= r*w

mgh1 = 0.5(2/5)m*r^2*W^2 + 0.5*m(r^2 *w^2)

mgh1 = 0.5*m*r^2*W^2 (2/5+1)

w^2= (5*g*h1)/(7*0.5*r^2)

W = sqrt((5*g*h1)/(7*0.5*r^2))

substituting the values

W = sqrt((5*9.8*0.83)/(7*0.5*0.027^2))

W = 126.25 rad/s

Part B

as thereis no friction so that there is no rotational force only trnaslational motion

so that the kinetic energy = gravitaional potential energy

0.5*m*v^2 = mgh2

v^2 = 2*g*h2

h2 = (r^2*W^2/2*g)

h2 = (0.027^2*126.25^2)/(2*9.8)

h2 = 0.593 m

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