Problem A rod of length L = 3.00 m and mass of M = 1.00 kg has uniform mass dens

Question

Problem A rod of length L = 3.00 m and mass of M = 1.00 kg has uniform mass density. It can rotate freely about its pivot at point O shown in the figure. The pivot is frictionless. 0 L/2 a) The rod is released from rest in a horizontal position as shown in the figure. What is the rod's angular speed when the tip of the rod reaches its lowest point? i.e. the rod rotates to a vertical position as shown in the figure. b) At this vertical position, what is the linear speed of the center of mass of the rod and the linear speed of the lowest point on the rod? c) Again, the rod is released from rest in a horizontal position as shown in the figure. What is the angular speed of the rod when it is at 45.0° to the horizontal? i.e. this angle is half of the 90.0° angle the rod rotated in part a).

Explanation / Answer

initial mechanical energy Ei = M*g*L/2

at the vertical position

final mechanical energy Ef = (1/2)*I*w^2

I = moment of inertia = (1/3)*1*3^2 = 3 kg m^2

from energy conservation

Ef = Ei

M*g*L/2 = (1/2)*I*w^2

M*g*L/2 = (1/2)*(1/3)*M*L^2*w^2

w = sqrt(3*g/L)

angular speed w(omega) = sqrt(3*9.8/3) = 3.13 rad/s

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part b)

linear speed v = r*w

linear speed of center of mass = (L/2)*w = (3/2)*3.13 = 4.695 m/s

linear speed of lowest point = (L)*w = (3)*3.13 = 9.39 m/s

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part c

initial mechanical energy Ei = M*g*L/2*(1-cos45)

at the vertical position

final mechanical energy Ef = (1/2)*I*w^2

I = moment of inertia = (1/3)*1*3^2 = 3 kg m^2

from energy conservation

Ef = Ei

M*g*L/2*(1-cos45) = (1/2)*I*w^2

M*g*L/2 = (1/2)*(1/3)*M*L^2*w^2

w = sqrt(3*g*(1-cos45)/L)

angular speed w(omega) = sqrt(3*9.8*(1-cos45)/3) = 1.694 rad/s

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