(8%) Problem 5: A thin cylindrical ring starts from rest at a height h, = 96 m.

Question

(8%) Problem 5: A thin cylindrical ring starts from rest at a height h, = 96 m. The ring has a radius R = 48 cm and a mass M = 7 kg. R: h, ©theexpertta.com e 20% Part a) Write an expression for the nng's initial energy at point l assuming that the gravitational potential energy at point 3 is zero. 20% Part (b) If the ring rolls (without slipping) all the way to point 2, what is the ring's energy at point 2 in terms of h2 and v2? 20% Part (c) Given h2 = 23 m, what is the velocity of the ring at point 2 in ms? 20% Part (d) What is the ring's rotational velocity in rad/s? X 20% Part (e) After passing point 2 the hill becomes frictionless and the ring's rotational velocity re velocity of the ring at point 3 in m/s? ains constant what is the linear Grade Summary v3 = 48.31

Explanation / Answer

using law of conservation of energy

Energy at point 1 = energy at point 3

m*g*h1 = 0.5*I*w^2 + 0.5*m*v^2

m*g*h1 = 0.5*I*w^2 + 0.5*m*v^2

I is the moment of inertia = m*R^2 = 7*0.48^2 = 1.6128 kg-m^2

w is the angular velocity at point 2

using law of conservation of energy

energy at 1 = energy at 2

m*g*h1 = (m*g*h2)+(0.5*I*w^2)+(0.5*m*v^2)

v = R*w

then

m*g*h1 = (m*g*h2)+(0.5*m*R^2*w^2)+(0.5*m*R^2*w^2)

m*g*h1 = (m*g*h2)+(m*R^2*w^2)

m cancels

9.81*96 = (9.81*23)+(0.48^2*w^2)

w = 55.75 rad/sec

Now

Energy at point 1 = energy at point 3

m*g*h1 = 0.5*I*w^2 + 0.5*m*v^2

m*g*h1 = 0.5*I*w^2 + 0.5*m*v^2

m*g*h1 = (0.5*m*R^2*w^2)+(0.5*m*v^2)

m cancels

9.81*96 = (0.5*0.48^2*55.75^2)+(0.5*v^2)

v = 34.16 m/sec

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