Tags: moment of inertia, angular momentum, torque, conservation of momentum, center of mass, kinetic energy, statics, beam reactions, ENGR 216, Texas A&M
This covers the mechanics topics tested in the ENGR 216 final: centre of mass, moment of inertia for various geometries, conservation of angular momentum, torque, rotational kinetic energy, conservation of linear momentum for collisions, and static equilibrium of beams. These appear in multiple-choice, short-answer, and long-answer sections.
Centre of mass
The mass-weighted average position of all particles in a system. Its location depends only on the masses and their positions, nothing else.
Moment of inertia (I)
The rotational equivalent of mass. Measures resistance to angular acceleration. For a system of particles: I = sum(m_i * r_i^2). The distance r_i is raised to the power of 2.
Angular momentum (L)
For a point mass: L = m v r (where r is the perpendicular distance from the rotation axis). For a rigid body: L = I * omega.
Conservation of angular momentum
When no external torque acts on a system, total angular momentum is constant: I1 omega1 = I2 omega2.
Torque (tau)
The rotational equivalent of force. Equals the rate of change of angular momentum: tau = delta-L / delta-t. Also tau = I * alpha.
Angular acceleration (alpha)
Rate of change of angular velocity: alpha = tau / I.
Linear momentum
p = m * v. Conserved in the absence of external forces.
Static equilibrium
A body is in static equilibrium when the sum of all forces and the sum of all moments (torques) about any point are both zero.
The location of the centre of mass depends only on the masses of the particles and their respective positions
It does not have to lie inside any physical object (think of a ring or a hollow shell)
It is not always at rest; it accelerates in response to net external force
The kinetic energy of the system is not maximised at the centre of mass
Common geometries tested:
Solid disk or cylinder about its central axis: I = (1/2) m r^2
Rectangular plate about an axis through its centre: I_cm = (1/12) m (a^2 + b^2)
Uniform rod or board pinned at one end: I_pin = (1/3) m L^2
Key point on the rectangular plate: if the rotation axis z passes through the centre of mass, then I_z / I_cm = 1. The ratio is 1 because the axis described is the centre-of-mass axis itself.
The distance r_i in the particle formula I = sum(m_i * r_i^2) is raised to the power of 2. This is a fill-in-the-blank exam point.
When moment of inertia changes (no external torque):
I1 omega1 = I2 omega2
Example: a person on a rotating platform moves bricks inward.
I decreases from 6.0 to 2.0 kg m^2
Initial omega = 1.2 rev/s
New omega = (6.0 * 1.2) / 2.0 = 3.6 rev/s
Kinetic energy ratio when I changes:
KE = (1/2) I omega^2
If I drops by a factor of 3, omega rises by a factor of 3 (from conservation of L)
New KE / Old KE = [(1/2)(I/3)(3*omega)^2] / [(1/2)(I)(omega^2)] = (1/3)(9) = 3
The ratio is exactly 3. Kinetic energy increases because the person does work pulling the bricks inward
Torque as rate of change of angular momentum:
tau = delta-L / delta-t
Example: flywheel with I = 0.140 kg m^2, angular momentum drops from 3.00 to 0.800 kg m^2/s in 1.50 s.
tau = (3.00 - 0.800) / 1.50 = 2.2 / 1.5 = 1.47 Nm
Torque from Newton's second law for rotation:
tau = I * alpha, so alpha = tau / I
Example: motor applies 30 Nm to a solid disk (m = 30 kg, r = 0.25 m), starting from rest, for 15 s.
I = (1/2)(30)(0.25^2) = 0.9375 kg m^2
alpha = 30 / 0.9375 = 32 rad/s^2
omega = 0 + 32 * 15 = 480 rad/s
Convert to rev/s: 480 / (2*pi) = 76.4 rev/s
For objects moving at perpendicular distance r from a point:
L = m v r for each object
If rotating in the same direction, angular momenta add directly
Example: Object 1 (6.0 kg, 2.0 m/s, r = 1.5 m), Object 2 (3.0 kg, 3.5 m/s, r = 2.5 m), same rotational direction.
L1 = 6.0 2.0 1.5 = 18.0 kg m^2/s
L2 = 3.0 3.5 2.5 = 26.25 kg m^2/s
L_total = 44.25 kg m^2/s
KE1 = 0.5 6.0 2.0^2 = 12 J
KE2 = 0.5 3.0 3.5^2 = 18.375 J
KE_total = 30.38 J
For a bullet passing through a block:
m_b v_b1 + M_w v_w1 = m_b v_b2 + M_w v_w2
Example: 5.20 g bullet at 672 m/s hits a 700 g block at rest, exits at 428 m/s.
Convert bullet mass: 0.00520 kg
(0.00520)(672) + 0 = (0.00520)(428) + (0.700)(v_w2)
3.4944 = 2.2256 + 0.700 * v_w2
v_w2 = 1.2688 / 0.700 = 1.81 m/s
For a simply supported beam with symmetric loading:
Calculate beam weight: W = m * g
Sum all downward forces at the centre
For a symmetric setup (supports at both ends, load at centre), each reaction = total downward force / 2
Example: 120 kg beam, 6 m long, supports at A (left) and B (right), 200 N force at centre.
Beam weight = 120 * 9.8 = 1176 N (acts at centre)
Total downward at centre = 1176 + 200 = 1376 N
R_A = R_B = 1376 / 2 = 688 N
Moment of inertia (particles): I = sum(m_i * r_i^2)
Solid disk: I = (1/2) m r^2
Rectangular plate at centre: I = (1/12) m (a^2 + b^2)
Rod pinned at end: I = (1/3) m L^2
Angular momentum: L = I omega, or L = m v * r
Conservation of angular momentum: I1 omega1 = I2 omega2
Torque: tau = delta-L / delta-t = I * alpha
Rotational kinetic energy: KE = (1/2) I omega^2
Linear momentum conservation: m1v1 + m2v2 = m1v1' + m2v2'
Equilibrium: sum(F) = 0, sum(M) = 0
Unit conversion: omega(rev/s) = omega(rad/s) / (2*pi)
⚠️ The centre of mass depends only on masses and positions. It does not have to lie inside any physical body.
⚠️ In the moment of inertia formula, r is raised to the power of 2. Tested as fill-in-the-blank.
⚠️ When I_z and I_cm use the same axis, their ratio is 1. The question is testing whether you recognise that the axis described is the centre-of-mass axis.
⚠️ The KE ratio when moment of inertia changes from 6.0 to 2.0 is exactly 3 (not 1/3). Energy is not conserved here because the person does work.
⚠️ Convert grams to kilograms in momentum problems. The bullet mass trap (5.20 g = 0.00520 kg) appears every year.
⚠️ Convert rad/s to rev/s by dividing by 2*pi, not by multiplying.
Q: A person on a rotating platform reduces their moment of inertia from 6.0 to 2.0 kg m^2. Initial angular speed is 1.2 rev/s. What is the new angular speed?
A: omega2 = (6.0 * 1.2) / 2.0 = 3.6 rev/s.
Q: What is the ratio of new KE to old KE when the moment of inertia drops from 6.0 to 2.0 kg m^2?
A: The ratio is exactly 3.
Q: A flywheel's angular momentum decreases from 3.00 to 0.800 kg m^2/s in 1.50 s. What is the average torque?
A: tau = (3.00 - 0.800) / 1.50 = 1.47 Nm.
Q: For a rectangular plate rotating about an axis through its centre, what is I_z / I_cm?
A: 1, because the axis described is the centre-of-mass axis.
Q: A 5.20 g bullet at 672 m/s passes through a 700 g block at rest and exits at 428 m/s. What is the block's speed?
A: 1.81 m/s, using conservation of linear momentum.
Q: True or false: the centre of mass depends only on the masses and positions of particles in the system.
A: True.
moment of inertia, rotational inertia, angular momentum, conservation of angular momentum, torque, angular acceleration, centre of mass, center of mass, kinetic energy, rotational kinetic energy, linear momentum, conservation of momentum, bullet-block collision, static equilibrium, beam reactions, simply supported beam, flywheel, solid disk, ENGR 216, Texas A&M