Today Ch 8 Rotational Motion Center of Gravity

Today: (Ch. 8) Rotational Motion

Center of Gravity • For the purposes of calculating the torque due to the gravitational force, you can assume all the force acts at a single location • The location is called the center of gravity of the object – The center of gravity and the center of mass of an object are usually the same point

Rotational Equilibrium • Equilibrium may include rotational equilibrium • An object can be in equilibrium with regard to both its translation and its rotational motion • Its linear acceleration must be zero and its angular acceleration must be zero • The total force being zero is not sufficient to ensure both accelerations are zero

Equilibrium Example • The applied forces are equal in magnitude, but opposite in direction • Therefore, ΣF = 0 • However, the object is not in equilibrium • The forces produce a net torque on the object • There will be an angular acceleration in the clockwise direction • For an object to be in complete equilibrium, the angular acceleration is required to be zero • Στ = 0 • This is a necessary condition for rotational equilibrium – All the torques will be considered to refer to a single axis of rotation – The same ideas can also be applied to multiple axes

Rotational Equilibrium, Lever • Use rotational equilibrium to find the force needed to just lift the rock – We can assume that the acceleration is zero – Also ignore the mass of the lever • The force exerted by the person can be less than the weight of the rock • The lever will amplify the force exerted by the person – If Lperson > Lrock

Tipping a Crate • We can calculate the force that will just cause the crate to tip • When on the verge of tipping, static equilibrium applies • If the person can exert about half the weight of the crate, it will tip

Moment of Inertia • The moment of inertia composed of many pieces of mass is • The moment of inertia of an object depends on its mass and on how this mass is distributed with respect to the rotation axis • The definition can be applied to find the moment of inertia of various objects for any rotational axis • Units of moment of inertia are kg · m 2

Various Moments of Inertia

Rotational Dynamics • Newton’s Second Law for a rotating system states Στ = Iα • Once the total torque and moment of inertia are found, the angular acceleration can be calculated • Then rotational motion equations can be applied • For constant angular acceleration:

Kinematic Relationships

Real Pulley with Mass, Example • Up to now, we have assumed a massless pulley • Using rotational dynamics, we can deal with real pulleys • The torque on the pulley is due to the tension in the rope • Apply Newton’s Second Laws for translational motion and for rotational motion • The crate undergoes translational motion • The pulley undergoes rotational motion • For the pulley: • The tension in the rope supplies the torque • The pulley rotates around its center, so that is a logical axis of rotation

Motion of a Crate, Example, cont • Pulley equation – Στ = - T R = Ipulley α – The pulley is a disc, so I = ½ mpulley R²pulley • For the crate – Take the +y direction as + – Equation: ΣF = T – mcrate g = mcrate a • Relating the accelerations – a = α Rpulley • Combine the equations and solve

Example If the mass of a wheel is increased by a factor of 17 and the radius is increased by a factor of 9, by what factor is the moment of inertia increased? A factor of

Example If the mass and height of the object is increased by twice, what would be the increase/decrease of final potential energy?

Example If final velocity is increased by 3 times, what would be increase/decrease in final kinetic energy?

Tomorrow: (Ch. 9) Energy and Momentum of Rotational Motion