this evaluates to approximately 27 radians. Using x = rθ , the distance traveled is closest to 14 m.5 . (a) The rotational inertia of the entire contraption is the sum of the moments of inertia for each part. I
for the rod is given; I for a point mass a distance L from the pivot is mL 2 . So, I (^) total = (1/3)ML 2 +
m L^2 . Be sure to differentiate between M and m .
(b) Rotational kinematics won’t work here because angular acceleration isn’t constant. We must use
energy.
U 1 + E 1 = U 2 + K 2
Define U = 0 at the bottom of the contraption when it hangs vertically. Then, U 2 is only caused by the
rod’s mass, which is concentrated L /2 above the zero point, so U 2 = MgL /2. U 1 is due to all of the
mass, concentrated L above the zero point: U 1 = (M + m )gL . K 1 = 0, and K 2 is unknown.
(M + m )gL + 0 = MgL /2 + ½ Iω )2.
Plug in I from part (a ) and solve for ω to get
(c) Just use v = rω . Here r = L because the center of rotation is L away from the mass.
(d) At this position, the mass is instantaneously in uniform circular motion. So, acceleration (and
therefore net force) must be centripetal. Net force is straight up, toward the center of rotation.
Rapid Review
• Rotational kinematics is very similar to linear kinematics. But instead of linear velocity, you work with
angular velocity (in radians/s); instead of linear acceleration, you work with angular acceleration (in
radians/s^2 ); and instead of linear displacement, you work with angular displacement (in radians).
• When doing rotation problems, work in radians, not degrees.
• Rotational inertia is the rotational equivalent of mass—it’s a measure of how difficult it is to start or
stop an object spinning.
• The rotational equivalent of Newton’s second law says that the NET torque on an object equals that