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Part 4: Rotational Motion & Oscillations
4.1: Rotational Motion
Rotational Kinematic Quantities (and Their Translational Analogs)
r = radius [m]
s = distance traveled [m]
v = tangential velocity [m/s]
at = tangential acceleration [m/s/s]
ar = radial (centripetal) acceleration [m/s/s]
Rotational Quantity / Symbol/ Definition / Units / Translational Analog / Relationship to Analogangle / q / [rad] / distance / q = s / r
angular velocity / w = dq / dt / [rad/s] / velocity / w = v / r
angular acceleration / a = dw / dt / [rad/s/s] / acceleration / a = at / r
Equations of Motion for Constant Angular Acceleration
Example:
A typical roulette wheel is initially spun at 18 radians per second and takes 10 seconds to stop. Its diameter is 80 cm. Find the following.
(a) initial tangential speed of a point on the rim of the wheel
(b) initial frequency of the wheel in revolutions per second
(c) the angular acceleration of the wheel assuming that the wheel stops uniformly
(d) the number of revolutions made by the wheel during its motion
Ans. (a) 7.2 m/s (b) 2.86 rps (c) -1.8 rad/s/s (d) 14.3
Example:
A wheel rolling down a hill speeds up uniformly from rest to 10 m/s in 20 seconds. The radius of the wheel is 50 cm.
(a) What is the angular acceleration of the wheel?
(b) What is the angular speed of the wheel after the 20 seconds has elapsed?
(c) How many revolutions does the wheel make during the 20 seconds?
(d) What is the centripetal acceleration of a section of the tread at this 20 second mark?
Ans. (a) 1 rad/s2 (b) 20 rad/s (c) 31.8 (d) 200 m/s2
Moment of Inertia
dm = mass of tiny chunk
r = distance from that chunk to the rotation axis
mks units [kg-m2]
The moment of inertia is the rotational analog of inertial mass. (It replaces m in equations.)
Moments of Inertia for Some Objects of Uniform Density
Rotated About A Center-of Mass Axis
Thin Rod / ICM = (1/12)ML2Hoop or Hollow Cylinder / ICM = MR2
Solid Cylinder or Disk / ICM = (1/2)MR2
Solid Sphere / ICM = (2/5)MR2
Parallel-Axis Theorem
I = ICM + MD2
Rotational Kinetic Energy
K = (1/2) I w2 mks units [J]
Example:
A 2-kg uniform brass cylinder with a 4-cm radius is rotated about its central axis at 30 revolutions per second. (a) Find its kinetic energy. (b) What would the cylinder’s kinetic energy be if its rotation axis was moved 2 cm from the center and it spun at this same rate?
Ans. (a) 28.4 J (b) 42.6 J
Torque
Size: t = r F sinq mks units [N-m]
Direction: Right Hand Rule – Fingers trace r into F, thumb gives t direction.
Torque is the rotational analog of force. (It replaces force F in equations.)
Newton’s Second Law for Rotations
Example:
A uniform, solid cylinder with a radius of 2 cm and a mass of 500 grams has a long piece of string wrapped around its circular edge. The string is pulled with a constant 0.75-N force for two seconds causing the cylinder to rotate about an axis that runs through its center. Find (a) the torque produced by the force, (b) the angular acceleration of the cylinder, (c) the kinetic energy of the cylinder at the end the 2 seconds, and (d) the tangential speed of a point on the edge of the cylinder at the end of the 2 seconds.
Ans. (a) 0.015 N-m (b) 150 rad/s2 (c) 4.5 J (c) 6 m/s
Static Equilibrium
There are two conditions if an object does not change its translational or rotational motion.
Net force must be zero.
Net torque must be zero.
Example:
A 2-kg board is balanced on a fulcrum. A 3-kg block is then placed 20 cm to the left of the fulcrum.
(a) Where should a 2-kg block be placed so that the board is once again balanced?
(b) How large is the normal force exerted by the fulcrum on the block?
(c) The blocks are removed from the board and the fulcrum is moved 15 cm to the right of its original position. Where should the 3-kg block be placed to once again balance the board?
Ans. (a) 30 cm to the right of the fulcrum (b) 68.6 N (c) 10 cm to the right of the fulcrum
Work and Power
Work done by torque: If torque is constant,
Power:
4.2: Rolling Motion
Rolling motion is a combination of translation and rotation.
Kinetic Energy of Rolling Object
ICM = moment of inertia about the center-of-mass axis
w = angular velocity = v / R
Rolling Down a Ramp
Acceleration: Speed at Bottom:
Rolling “Yo-Yos”
As done in class, know how to apply Newton’s Second Law for Translations and Rotations to the different geometries.
4.3: Angular Momentum
An object is rotating about an axis at angular speed w. The object has a moment of inertia I for this axis. The object has an angular momentum L where
L = Iw mks units [kg-m2/s = J-s]
Newton’s Second law for Rotations (Revisited)
Conservation of Angular Momentum
Single Rotating Object: Iiwi = Ifwf
Example:
An ice skater is doing a spin at 2 revolutions per second with her arms held out from her body. She brings her arms in, reducing her moment of inertia to 0.5 kg-m2 and increasing her spin rate to 10 revolutions per second.
(a) Find the skater’s initial moment of inertia when her arms were held out.
(b) Find the size of the skater’s angular momentum.
Ans. (a) 2.5 kg-m2 (b) 31.4 J
4.4: Simple Harmonic Motion
Mass-on-a-Spring
An object with mass m is connected to a spring with spring constant k and equilibrium position xo = 0. The object is pulled to position x = A and released.
Natural oscillation frequency: [ rad/s ]
[ cycles/s = Hz ]
Natural period: [ s ]
Position:
Velocity: Maximum speed: vmax = wA
Acceleration: Maximum acceleration: amax = w 2 A
Total Energy: ETOT =K + U = (1/2)kA2 constant
Simple Pendulum
The bob of a pendulum with mass m and length L is pulled to a small starting angle of qo with respect to the vertical and released.
Natural oscillation frequency: [ rad/s ]
[ cycles/s = Hz ]
Natural period: [ s ]