(690) Problem 16: A thin cylindrical ring starts from rest at a height h,-75 m.

Question

(690) Problem 16: A thin cylindrical ring starts from rest at a height h,-75 m. The ring has a radius R-8 cm and a mass M= 4 kg R: Otheexpertta.com 20% Part (a) Write an expression for the rings initial energy at point l assuming that the gravitational potential energy at point 3 1s zero O 20% Part (b) If the ring rolls (without slipping) all the way to point 2, what is the ring's energy at port 2 n terms of h2 and v2? O 20% Part (c) Given h,-29 m, what is the velocity of the ring at point 2 in m s? 20% Part (d) What is the ring's rotational velocity in rad s? * 20% Part (e) After passing point 2 the hill becomes frictionless and the ring's rotational velocity remains constant. What is the linear velocity of the ring at point 3 in m/s? Grade Summa Deductions Potential ry V321.42 0% 100% tanO ( 789 HOME Submissions Attempts remaining:;3 detailed view Sih cos cotanasin acos0 atan acotan sinhO cosh0 tan0 cotanh0 Degrees Radians (000 per attempt) 0% 0% 0% 0% END N BACKSPACE DEL CLEAR Submit Hint I give up!

Explanation / Answer

initial energy E1 = M*g*h1

part(b)

at point 2 energy E2 = M*g*h2 + (1/2)*I*w2^2 + (1/2)*M*v2^2

I = M*R^2

w2 = v2/R2

E2= M*g*h2 + (1/2)*M*R^2*(V2/R2)^2 + (1/2)*M*v2^2

E2 = M*g*h2 + (1/2)*M*v2^2 + (1/2)*M*v2^2 = M*g*h2 + M*v2^2

part(c)

E2 = E1

M*g*h2 + M*v2^2 = M*g*h1

g*h2 + v2^2 = g*h1

v2 = sqrt(g*(h1-h2))

v2 = sqrt(9.8*(75-29)) = 21.23 m/s

part(d)

w2 = v2/R2 = 21.23/0.08 = 265.4 rad/s

part(e)

total energy at point 3

E3 = (1/2)*M*v3^2

E2 = E3

M*g*h1 = (1/2)*M*v3^2

v3 = sqrt(2*g*h1)

v3 = sqrt(2*9.8*75)

v3 = 38.34 m/s

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