Question

A stick is released from rest when it is perfectly horizontal. Assume its pivot is

frictionless. Use conservation of energy to nd an expression for the angular velocity

of the stick when it reaches its vertical position. (Use: Irod = 1/3*mL^2. Your answer

should be in terms of g and L.) For a stick of length 2.0 m, what would be the nal

angular velocity?

A stick is released from rest when it is perfectly horizontal. Assume its pivot is frictionless. Use conservation of energy to find an expression for the angular velocity of the stick when it reaches its vertical position. (Use: Irod-3mL2. Yo! should be in terms of g and L.) For a stick of length 2.0 m, what would be the final angular velocity? L'2. Your answer 「. Your answer Center of Mass

Answer 1

Rod in vertical position:

Gravitational potential energy, Ui=MgL/2

As it is in equilibrium and not moving kinetic energy, Ki=0

Total mechanical energy, Ei=Ui+Ki = MgL/2 -----------(1)

Rod in horizontal position:

As the rod falls and just reaches the horizontal position, its centre of mass is in the ground. Therefore gravitational potential energy, Uf = 0

As it is in equilibrium and not moving kinetic energy, Kf=(1/2)Iω²

where I is the moment of inertia about the axis of rotation which is the point of contact with the ground.

Total mechanical energy, Ef = Uf+Kf =(1/2)Iω²------------(2)

As gravitational force is conservative in natutre, the total mechanical enrgy remains constant.

Therefore, E_f = Ei

Equating (2) and (1), we get, (1/2)I*ω²=MgL/2

As we need angular velocity about the point of conatct, we determine moment of inertia also about the same point.

So, moment of inertia I=ML² /3

Using this we get, (1/2)*(ML²/3)*ω² = MgL²

Or, (L²/3)*ω² = gL

Which gives, (L/3)ω² = gL

Or, ω²=3g / L

The angular speed, ω = √(3g/L)

ω = √(3*9.8 / 2.0m) = 3.83 rad/s