an object m = 3.05kg rolls from rest down an incline plane, its center initially
ID: 2034703 • Letter: A
Question
an object m = 3.05kg rolls from rest down an incline plane, its center initially at a height of 7.55m above the bottom of the ramp. The inclination of the incline as measured from the horizontal is 37.3o. If the object is a solid uniform cylinder of radius r = 1.25m, then at the bottom of the incline, what is the (a) rotational kinetic energy and (b) the angular momentum. If the object is a hoop of radius, r = 1.05m, then at the bottom of the ramp, what is the (c) rotational kinetic energy and (d) the angular momentum?
Explanation / Answer
a) Unifrom cyllinder
Let w is the angular speed and v is the linear speed at the bottom.
Apply conservation of energy
(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h
(1/2)*m*v^2 + (1/2)*(1/2)*m*r^2*w^2 = m*g*h
(1/2)*m*v^2 + (1/4)*m*(r*w)^2 = m*g*h
(1/2)*m*v^2 + (1/4)*m*v^2 = m*g*h
(3/4)*m*v^2 = m*g*h
v = sqrt(4*g*h/3)
w = v/r
= sqrt(4*g*h/3)/r
= sqrt(4*9.8*7.55/3)/1.25
= 7.94 rad/s
KE_rotationa = (1/2)*I*w^2
= (1/2)*(1/2)*m*r^2*w^2
= (1/4)*3.05*1.25^2*7.94^2
= 75.1 J
b) angular momentum = I*w
= (1/2)*m*r^2*w
= (1/2)*3.05*1.25^2*7.94
= 18.9 kg.m^2/s
c) Hoop
Let w is the angular speed and v is the linear speed at the bottom.
Apply conservation of energy
(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h
(1/2)*m*v^2 + (1/2)*m*r^2*w^2 = m*g*h
(1/2)*m*v^2 + (1/2)*m*(r*w)^2 = m*g*h
(1/2)*m*v^2 + (1/2)*m*v^2 = m*g*h
m*v^2 = m*g*h
v = sqrt(g*h)
w = v/r
= sqrt(g*h)/r
= sqrt(9.8*7.55)/1.05
= 8.19 rad/s
KE_rotationa = (1/2)*I*w^2
= (1/2)*m*r^2*w^2
= (1/2)*3.05*1.05^2*8.19^2
= 113 J
d) angular momentum = I*w
= m*r^2*w
= 3.05*1.05^2*8.19
= 27.5 kg.m^2/s
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