Question

two cylinders with the radius r= 0.650 m are rolled with out slipping down an incline that desends a vertical distance of 2.45 meters each cylinder has equal mass m=3.68kg but one is solid the other is hollow shell

a) what is the center mass velocity of the solid cylinder at the bottom of the incline

b) what is the center mass velocity of the holllow cylinder of the botom of the incline

c) what is the angular frequency of each cylinder when it reaches the bottom of the incline

Answer 1

a) for solid cyllinder:

let v is the linear speed and w is the angular speed at the bottom

Apply conservation of momentum

final kinetic energy = initial potential 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 (since v = R*w)

(3/4)m*v^2 = m*g*h

v_solidcyllinder = sqrt(4*g*h/3)

= sqrt(4*9.8*2.45/3)

= 5.66 m/s <<<<<<<<<---------------Answer

b) for hallow cyllinder:
let v is the linear speed and w is the angular speed at the bottom

Apply conservation of momentum

final kinetic energy = initial potential 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 (since v = R*w)

m*v^2 = m*g*h

v_hollowcyllinder = sqrt(g*h)

= sqrt(9.8*2.45)

= 4.90 m/s <<<<<<<<<---------------Answer

c) angular frequency of solid cyllinder,
w_solidcyllinder = v_solidcyllinder/R

= 5.66/0.65

= 8.71 rad/s <<<<<<<<<---------------Answer

angular frequency of hollow cyllinder,
w_hollowcyllinder = v_hollowcyllinder/R

= 4.90/0.65

= 7.54 rad/s <<<<<<<<<---------------Answer