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?

Answer 1

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