Torque is equal to the force times the perpendicular distance from the axis of rotation to the point
of application of the force, or the distance from the axis times the component of the force
perpendicular to the force.sinr F rF rF
⊥ ⊥= = =
[1] Newton’s second law for rotation isI
= and the linear version, written here for tangential
variables, istan tannetF ma= .
Determine the units for the moment of inertia, I, by usingtan netrF
= in the rotational second law.
[2] A mass m is tied to a string which is wrapped around a frictionless pulley of mass M and
radius R. The mass and radius of the string are negligible.
a) Set up the Newton’s second law equation for the hanging
mass.
b) Set up the rotational Newton’s second law for the pulley.
Use I = ½ mr2
c) Combine these to solve for the linear acceleration.
[3] A thin rod of mass M=0.2[kg] and length L has a 0.1[kg] mass
hanging from it at L/4 from the left end and two upward forces acting on it: F1 at L/8 from the
right end and F2 at L/3 from the left end. Determine the values of F1 and F2 so that the system is
in equilibrium (0a
= = ).y
x
0
0
0
net
net
net
F
F
= =
= =
= =
You can choose any point as the axis of rotation in
equilibrium.