You are consulting for an amusement park that wants to build a new \"Rotor\" rid

Question

You are consulting for an amusement park that wants to build a new "Rotor" ride. In order to increase capacity, they would like to build a unit with a 15-ft diameter. The Rotor should provide a centripetal acceleration of 3g. What must be the angular speed in rpm?

A solid sphere rolls on a level surface at a speed of 5.73 m/s. What is the ratio of its rotational kinetic energy to its total kinetic energy?

A spherical object with a moment of inertia of 0.481mr2 starts from rest rolling down a 2.81-m high incline. If the sphere is rolling without slipping, what is its linear speed at the bottom of the incline?

A spherical object with moment of inertia 0.6mr2 rolls without slipping down an incline. At the bottom of the incline, what fraction of its total kinetic energy is rotational kinetic energy? (Answer in percent; use % as units.

Please show work

Explanation / Answer

1)

d = 15 ft = 15*0.3048 = 4.572 m

r = d/2 = 2.286 m

a_rad = r*w^2

w = sqrt(a_rad/r) = sqrt(3*9.8/2.286) = 3.586 rad/s = 34.26 rpm

2)


KE_total = 0.5*m*v^2 + 0.5*I*w^2

KE_total = 0.5*m*v^2 + 0.5*(2/5)*m*r^2*w^2

KE_total = 0.7*m*v^2

KE_rotational = 0.2*m*v^2

KE_total/KE_rotational = 0.7/0.2 = 3.5

3)

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

m*g*h = 0.5*m*v^2 + 0.5*0.481*m*r^2*w^2

m*g*h = 0.981*m*v^2

v = sqrt(g*h/0.981) = sqrt(9.8*2.81/0.981) = 6.17 m/s

4)

KE_total = 0.8*m*v^2

KE_rotational = 0.3*m*v^2

KE_total/KE_rotational = 0.8/0.3 = 2.67

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