# Comparing rotational and linear motion

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Comparing rotational and linear motion

Angular Velocity Angular velocity,w, is the rate of change in angular displacement. (radians per second.) w = Angular velocity in rad/s. q t Angular velocity can also be given as the frequency of revolution, f (rev/s or rpm): w = 2pf Angular frequency f (rev/s).

Angular Acceleration Angular acceleration is the rate of change in angular velocity. (Radians per sec per sec.) The angular acceleration can also be found from the change in frequency, as follows:

Force and Linear Acceleration
When an object is subject to a net force, it undergoes an acceleration. (Newton’s 2nd) Torque and Angular Acceleration When a rigid object is subject to a net torque, it undergoes an angular acceleration.

Inertia of Rotation t Force does for translation what
Consider Newton’s second law for the inertia of rotation Linear Inertia, m = F/a m = = 5 kg 20 N 4 m/s2 F = 20 N a = 4 m/s2 Rotational Inertia, I I = = = 2.5 kg m2 (20 N)(0.5 m) 2 rad/s2 t a F = 20 N R = 0.5 m a = 2 rad/s2 Force does for translation what torque does for rotation:

Moment of Inertia This mass analog is called the moment of inertia, I, of the object is defined relative to rotation axis SI units are kg m2

I depends on both the mass and its distribution. If an object’s mass is distributed further from the axis of rotation, the moment of inertia will be larger.

Common Moments of Inertia
Common moments of inertia are on page 251.

Example 1 A circular hoop and a disk each have a mass of 3 kg and a radius of 20 cm. Compare their rotational inertias. R I = mR 2 Hoop I = kg m2 R I = ½mR 2 Disk I = kg m2

Important Analogies t m I x f
For many problems involving rotation, there is an analogy to be drawn from linear motion. R 4 kg w t wo = 50 rad/s t = 40 N m m x f I A resultant torque t produces angular acceleration a of disk with rotational inertia I. A resultant force F produces negative acceleration a for a mass m.

Example 2 Treat the spindle as a solid cylinder.
a) What is the moment of Inertia of the spindle? b) If the tension in the rope is 10 N, what is the angular acceleration of the wheel? c) What is the acceleration of the bucket? d) What is the mass of the bucket? M

Solution a) What is the moment of Inertia of the spindle?
Given: M = 5 kg, R = 0.6 m M = 0.9 kgm2

Solution b) If the tension in the rope is 10 N, what is a?
Given: I = 0.9 kg m2, T = 10 N, r = 0.6 m a = (0.6m)(10 N)/(0.9 kg∙m2) a = 6.67 rad/s2 c) What is the acceleration of the bucket? Given: r=0.6 m, a = 6.67 rad/s M a = (6.67 rad/s2)(0.6 m) a=4 m/s2

Solution d) What is the mass of the bucket?
Given: T = 10 N, a = 4 m/s2 M = 1.72 kg M

Comparing rotational and linear motion

Combined Rotation and Translation
vcm First consider a disk sliding without friction. The velocity of any part is equal to velocity vcm of the center of mass. v R P Now consider a ball rolling without slipping. The angular velocity  about the point P is same as  for disk, so that we write: Or

Two Kinds of Kinetic Energy
Kinetic Energy of Translation: K = ½mv2 v R P Kinetic Energy of Rotation: K = ½I2 Total Kinetic Energy of a Rolling Object: KE due to rotation KE of center-of-mass motion

Example 3 What is the kinetic energy of the Earth due to the daily rotation? Given: Mearth=5.98 x1024 kg, Rearth = 6.63 x106 m. First, find w = 7.27 x10-5 rad/s = 2.78 x1029 J

Summary – Rotational Analogies
Quantity Linear Rotational Displacement Displacement x Radians  Inertia Mass (kg) I (kgm2) Force Newtons N Torque N·m Velocity v “ m/s ”  Rad/s Acceleration a “ m/s2 ”  Rad/s2 Momentum mv (kg m/s) I (kgm2rad/s)

Analogous Formulas F = ma  = I K = ½mv2 K = ½I2 Work = Fx Work = tq
Linear Motion Rotational Motion F = ma  = I K = ½mv2 K = ½I2 Work = Fx Work = tq Power = Fv Power = I Fx = ½mvf2 - ½mvo2  = ½If2 - ½Io2

Example 4 A solid sphere rolls down a hill of height 40 m.
What is the velocity of the ball when it reaches the bottom? (Note: We don’t know r or m!) v = 23.7 m/s

Conserved if no net outside torques
Angular Momentum Rigid body Point particle Analogy between L and p Angular Momentum Linear momentum L = Iw p = mv t = DL/Dt F = Dp/Dt Conserved if no net outside torques Conserved if no net outside forces

Example 5 A 65-kg student sprints at 8.0 m/s and leaps onto a 110-kg merry-go-round of radius 1.6 m. Treating the merry-go-round as a uniform cylinder, find the resulting angular velocity. Assume the student lands on the merry-go-round while moving tangentially.

Solution Known: M, R, m, v0 Find: wF First, find L0 Next, find Itot
Now, given Itot and L0, find w = rad/s

= Summary of Formulas: Conservation: I = SmR2 mgho mghf Height? ½Iwo2
½mvo2 = mghf ½Iwf2 ½mvf2 Height? Rotation? velocity?