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## Dynamics of a Rigid Body

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**General Definition of Torque**• Let F be a force acting on an object, and let r be a position vector from a rotational center to the point of application of the force. The magnitude of the torque is given by • = 0° or = 180 °: torque are equal to zero • = 90° or = 270 °:magnitude of torque attain to the maximum**Torque Units and Direction**• The SI units of torque are N.m • Torque is a vector quantity • Torque magnitude is given by • Torque will have direction • If the turning tendency of the force is counterclockwise, the torque will be positive • If the turning tendency is clockwise, the torque will be negative The work done by the torque is given by**Net Torque**• The force will tend to cause a counterclockwise rotation about O • The force will tend to cause a clockwise rotation about O • St = t1 + t2 = F1d1 – F2d2 • If St 0, starts rotating • If St= 0, rotation rate does not change Rate of rotation of an object does not change, unless the object is acted on by a net torque**Power delivered by torque**• To find the instantaneous power delivered by the torque, divide both sides by dt or**Newton’s Second Law for a Rotating Object**• When a rigid object is subject to a net torque (≠0), it undergoes an angular acceleration • The angular acceleration is directly proportional to the net torque • The angular acceleration is inversely proportional to the moment of inertia of the object • The relationship is analogous to**Extended Work-Energy Theorem**• The work-energy theorem tells us • When Wnc = 0, • The total mechanical energy is conserved and remains the same at all times • Remember, this is for conservative forces, no dissipative forces such as friction can be present**Total Energy of a System**• A ball is rolling down a ramp • Described by three types of energy • Gravitational potential energy • Translational kinetic energy • Rotational kinetic energy • Total energy of a system**Conservation of Mechanical Energy**• Conservation of Mechanical Energy • Remember, this is for conservative forces, no dissipative forces such as friction can be present**Work-Energy in a Rotating System**• The work done on the body by the external torque equals the change in the rotational kinetic energy • The work equals the negative of the change in potential energy • Conservation of Energy in Rotational Motion**Problem Solving Hints**• Choose two points of interest • One where all the necessary information is given • The other where information is desired • Identify the conservative and non-conservative forces • Write the general equation for the Work-Energy theorem if there are non-conservative forces • Use Conservation of Energy if there are no non-conservative forces • Use v = w to combine terms • Solve for the unknown**Example**• A meterstick is initially standing vertically on the floor. If the falls over, with what angular velocity will it hi the floor? Moment of inertia is Ml2/2 l = 1.0 m U = mgy y0 = com = .5 m 0= 0 y0 = com = 0 m Therefore, What rate is gravity delivering energy? The mass of the meterstick is 0.15kg R= l/2**General Problem Solving Hints**• The same basic techniques that were used in linear motion can be applied to rotational motion. • F becomes • m becomes I • a becomes • v becomes ω • x becomes θ**Angular Momentum**• Similarly to the relationship between force and momentum in a linear system, we can show the relationship between torque and angular momentum • Linear momentum is defined as • Angular momentum is defined as**Angular Momentum and Torque**• Net torque acting on an object is equal to the time rate of change of the object’s angular momentum • Angular momentum is defined as • Analog in impulse**Angular Momentum Conservation**• If the net torque is zero, the angular momentum remains constant • Conservation of Angular Momentum states: The angular momentum of a system is conserved when the net external torque acting on the systems is zero. • That is, when • then**Blocks and Pulley**• Two blocks with masses m1 = 5 kg and m2 = 7 kg are attached by a string over a pulley with mass M = 2kg. The pulley, which runs on a frictionless axle, is a hollow cylinder with radius 0.05 m over which the string moves without slipping.**Blocks and Pulley**• Two blocks with masses m1 = 5 kg and m2 = 7 kg are attached by a string over a pulley with mass M = 2kg. The pulley, which runs on a frictionless axle, is a hollow cylinder with radius 0.05 m over which the string moves without slipping.**An automobile with rear-wheel drive is accelerating at**4.0m/s2along a straight road. Consider one of the wheels of this automobile. The axle pushes forward, providing an acceleration of 4.0 m/s2. Simultaneously, the friction force of the road pushes the bottom of the wheel backward, providing a torque that gives the wheel an angular acceleration. The wheel has a radius of 0.38m and a mass of 25kg. Find the backward force that the friction force exerts on the wheel, and find the forward force that the axle exerts on the wheel.