4.1 The Concepts of Force and Mass
... PRINCIPLE OF CONSERVATION OFANGULAR MOMENTUM The angular momentum of a system remains constant (is conserved) if the net external torque acting on the system is zero. ...
... PRINCIPLE OF CONSERVATION OFANGULAR MOMENTUM The angular momentum of a system remains constant (is conserved) if the net external torque acting on the system is zero. ...
Chapter 8 Rotational Dynamics conclusion
... your right hand, so that your fingers circle the axis in the same sense as the rotation. ...
... your right hand, so that your fingers circle the axis in the same sense as the rotation. ...
POP4e: Ch. 1 Problems
... 13. A mouse is initially at rest on a horizontal turntable mounted on a frictionless, vertical axle. As the mouse begins to walk clockwise around the perimeter, which of the following statements must be true of the turntable? (a) It also turns clockwise. (b) It turns counterclockwise with the same a ...
... 13. A mouse is initially at rest on a horizontal turntable mounted on a frictionless, vertical axle. As the mouse begins to walk clockwise around the perimeter, which of the following statements must be true of the turntable? (a) It also turns clockwise. (b) It turns counterclockwise with the same a ...
Document
... We write an equation like this for every particle in the body and then add all these equations: ...
... We write an equation like this for every particle in the body and then add all these equations: ...
Here - UCSB HEP
... The moment of inertia changes because the distribution of mass around the axis changes. The angular momentum is constant because there is no external torque. Thus L = Ioldωold = Inewωnew Æ ωnew = (Iold/Inew) ωold Kold = ½ Iold ω2old Knew = ½ Inew ω2new = ½ Inew (Iold/Inew)2 ω2old = ½ (I2old/Inew) ω2 ...
... The moment of inertia changes because the distribution of mass around the axis changes. The angular momentum is constant because there is no external torque. Thus L = Ioldωold = Inewωnew Æ ωnew = (Iold/Inew) ωold Kold = ½ Iold ω2old Knew = ½ Inew ω2new = ½ Inew (Iold/Inew)2 ω2old = ½ (I2old/Inew) ω2 ...
PY1052 Problem Set 8 – Autumn 2004 Solutions
... It may seem odd, but the best way to go about this is to start at the top and work our way down. Looking at the top brick, in order for it to be in equilibrium, there must be no net force and no net torque acting on it. If we imagine starting with the top brick so that its right side is flush with t ...
... It may seem odd, but the best way to go about this is to start at the top and work our way down. Looking at the top brick, in order for it to be in equilibrium, there must be no net force and no net torque acting on it. If we imagine starting with the top brick so that its right side is flush with t ...
Final 1
... 17. A 10 kg sphere is glued to a massless stick that is tangent to it and then spun about the axis formed by the stick. What is the sphere's rotational inertia I about this axis, if its radius is 0.2 m ? The rotational inertia of a sphere about its center is Icm = 2/5 MR2 . A. 0. 24 kg.m2 B. 0.56 kg ...
... 17. A 10 kg sphere is glued to a massless stick that is tangent to it and then spun about the axis formed by the stick. What is the sphere's rotational inertia I about this axis, if its radius is 0.2 m ? The rotational inertia of a sphere about its center is Icm = 2/5 MR2 . A. 0. 24 kg.m2 B. 0.56 kg ...
