Chapter 6 Rotational File
... Center of Gravity and Center of Mass • The three equations giving the coordinates of the center of gravity of an object are identical to the equations giving the coordinates of the center of mass of the object • The center of gravity and the center of mass of the object are the same if the value of ...
... Center of Gravity and Center of Mass • The three equations giving the coordinates of the center of gravity of an object are identical to the equations giving the coordinates of the center of mass of the object • The center of gravity and the center of mass of the object are the same if the value of ...
Lecture #1, June 9
... We have also learned that the special steps have to be taken in order to add vectors. Both magnitude and direction have to be taken into account when adding, so vector addition can not be reduced to a simple algebraic addition. Today we shall talk about multiplying vectors. Once again vector multipl ...
... We have also learned that the special steps have to be taken in order to add vectors. Both magnitude and direction have to be taken into account when adding, so vector addition can not be reduced to a simple algebraic addition. Today we shall talk about multiplying vectors. Once again vector multipl ...
A solid disk with mass = 0
... 3) A figure skater with an initial moment of inertia of 40 kg.m2 spins at a rotational speed of 180 rpm. Assume there is no friction acting on the skater. a) What is the angular velocity of the skater (in SI units)? ...
... 3) A figure skater with an initial moment of inertia of 40 kg.m2 spins at a rotational speed of 180 rpm. Assume there is no friction acting on the skater. a) What is the angular velocity of the skater (in SI units)? ...
Chapter 8, Part V
... Displacement x θ Velocity v ω Acceleration a α Force (Torque) F τ Mass (moment of inertia) m I Newton’s 2nd Law ∑F = ma ∑τ = Iα Kinetic Energy (KE) (½)mv2 (½)Iω2 CONNECTIONS v = rω, atan= rα, aR = (v2/r) = ω2 r τ = rF, I = ∑(mr2) ...
... Displacement x θ Velocity v ω Acceleration a α Force (Torque) F τ Mass (moment of inertia) m I Newton’s 2nd Law ∑F = ma ∑τ = Iα Kinetic Energy (KE) (½)mv2 (½)Iω2 CONNECTIONS v = rω, atan= rα, aR = (v2/r) = ω2 r τ = rF, I = ∑(mr2) ...
Lesson 18 (1) Force on a Current Loop in a Uniform Magnetic field
... This energy is calculated from the work required to rotate the current loop. For that we first need the expression for the work done by a torque on an object. For simplicity, let the object be a rod of length L , and a force F acts perpendicularly to the rod at one end, so that the torque about the ...
... This energy is calculated from the work required to rotate the current loop. For that we first need the expression for the work done by a torque on an object. For simplicity, let the object be a rod of length L , and a force F acts perpendicularly to the rod at one end, so that the torque about the ...
