1. Motion IV Rotary Motion

2. Moment/Torque The moment or torque is the tendency of a force to rotate a body about a specified axis or point. Moment is denoted by the vector M Torque is denoted by the vector τ

3. M o = r x F τ o = r x F F r o

4. The vector r is any vector from the point o to any point on the line of action of the force F. The cross product is a vector that is perpendicular to both of the vectors making up the cross product, i.e. – The moment or torque vector is perpendicular to both r and F

5. Newton’s Second Law For translation, F = ma The net external force F acting on a body of mass m imparts an acceleration a. For rotation, τ = Iα The net external torque or moment τ acting on a body whose mass moment of inertia is I imparts an angular acceleration α.

6. Definitions The angular position is given by the vector ϴ – It must be expressed in radians (not degrees). The angular velocity is given by the vector ω – It must be expressed in radians/sec The angular acceleration is expressed by the vector α – It must be expressed in radians/sec 2

7. ω = dϴ/dt The angular velocity is the derivative with respect to time of the angular position. Radians/sec α = dω/dt The angular acceleration is the derivative with respect to time of the angular velocity. Radians/sec 2

8. Equations of Motion There are basically three equations of rotary motion similar to the three equations of translational motion. ϴ = ϴ o + ω o t + (1/2)αt 2 ω = ω o + αt ω = ω o 2 + 2α(ϴ - ϴ o )

9. Tangential Velocity The tangential velocity is equal to the product of the radius times the angular velocity, i.e. v t = rω r v t

10. Class Activity In ancient times, vinyl records (diameter 10”) were the medium for recording music. Long playing records turned at a constant angular velocity of 33-1/3 rpm (revolutions per minute). What is the tangential velocity of a point on the outer edge? You have been listening to one of these recordings when you realize that you are late for class. You lift the arm off the record and the motor for the turntable stops. Due to the low friction bearings used in making the turntable, the turntable does not stop rotating instantly, but ‘coasts’ to a stop. If the turntable coasts for 12.5 complete revolutions before stopping, calculate the constant angular deceleration and the duration (time) of the ‘coasting’.

11. Mass Moment of Inertia The mass moment of inertia I is a mass property of a body that measures the resistance of the body to rotation, i.e. τ = Iα The mass moment of inertia I is similar in rotation to the mass m in translation, i.e. F = ma

12. Consider a point mass at a distance of r from an axis of rotation, by definition, I = mr 2 m r axis of rotation

13. If the system undergoing rotation consists of several point masses m i at a distance of r i from the axis of rotation, then I = Σ r i 2 m i or more precisely, I = ∫r 2 dm

14. The moment of inertia of an object can change if its shape changes. Figure skaters who begin a spin with arms outstretched provide a striking example. By pulling in their arms, they reduce their moment of inertia, causing them to spin faster by the conservation of angular momentum.angular momentum

15. Mass Moments of Inertia for Shapes Thin solid disk of radius R and mass m rotating about an axis perpendicular to the disk at its center I = (1/2) mR 2 Solid cylinder of radius R, mass m about its centerline I = (1/2) mR 2

16. Solid sphere of radius R, mass m rotating about a diameter. I = (2/5)mR 2 Rod of length L and mass m rotating about an axis normal to the rod at its center I = (1/12)mL 2 Rod of length L and mass m rotating about an axis normal to the rod and at one end. I = (1/3)mL 2

17. Integration to Find I Determine the mass moment of inertia of a slender rod of length L and mass m with respect to an axis that is perpendicular to the rod and passes through one end of the rod. Choose a differential element between x and x + dx along the length of the rod. dm = (m/L)dx I = ∫x 2 dm = ∫x 2 (m/L)dx = (m/L) ∫x 2 dx I = (1/3)mL 2

18. Class Activity Using integration find the mass moment of I inertia of a slender rod of length L and mass m with respect to an axis perpendicular to the rod at its center Set x= 0 at the axis of rotation, i.e. the center of the length of the rod Choose a differential element between x and x + dx along the length of the rod. dm = (m/L)dx I = ∫x 2 dm = ∫x 2 (m/L)dx = (m/L) ∫x 2 dx With the limits of integration from x = –L/2 to x = +L/2

19. I = (m/L) (1/3) x 3 evaluated from –L/2 to +L/2. I = (m/L)(1/3)[(L 3 /8) – (- L 3 /8)] I = (m/L)(1/3)(L 3 /4) I = (1/12)mL 2

20. Units of I The dimensionality of I can be found [I] = [m][r 2 ] [I] = ML 2 The units of I would then be Kg m 2 in the SI system Slug ft 2 in the US system