Title: Physics 1710 Chapter 12 Rolling Motion and Angular Momentum
1Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Quiz
- What torque must be applied to a discus (assume
a short cylindrical shape) that has a diameter
of 20 cm and a mass of 3.0 kg, if it leaves the
hand of the discus thrower spinning at a rate of
180 rpm if he takes 0.5 sec to throw the disc?
? 2??(180)/60 18.7 rad/s I ½ MR2 ½ (3.0)
(0.10)2 0.015 kg m
2Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Quiz
- What torque must be applied to a discus (assume
a short cylindrical shape) that has a diameter
of 20 cm and a mass of 3.0 kg, if it leaves the
hand of the discus thrower spinning at a rate of
180 rpm if he takes 0.5 sec to throw the disc?
? 18.7 rad/s I ½ MR2 0.015 kg m ? 18.7
rad/s/(0.5 sec) 37.4 rad/s/s T I? (0.015)
(37.4) N?m 0.56 N?m F T/R 0.56/0.1 5.6 N
3Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- 1' Lesson
- The total Kinetic energy of a rotating system is
the sum of the rotational energy about the Center
of Mass and the translational KE of the CM. - T r x F
- Angular momentum L is the vector product of the
moment arm and the linear momentum. - The net externally applied torque is equal to
the time rate of change in the angular momentum.
4Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- 1' Lesson (contd)
- Angular momentum about an axis z is equal to
the product of the moment of inertia of the body
about that axis and the angular velocity about z. - In the absence of torques, the angular momentum
is conserved.
5Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Rolling Motion
- The compound motion of a rolling object is a
translation plus a rotation. - vCM R? aCM R?
- Cylinder rolling down a ramp
- F Mg sin ? T RF I ?
- aCM R 2 (Mg sin ?)/I x ½ aCM t 2
- Which will win? A large disk or a small one?
- A heavy one or a light one?
6Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Total Energy of Motion
- K ½ Ip ? 2 ½ ICM ? 2 ½ MR 2 ? 2
- By parallel axis theorem.
- Thus, the total kinetic energy is the energy of
motion of the center of mass plus the energy of
rotation about the CM
7Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Vector Product
- C A x B
- Cx Ay Bz Az By
- Cyclically permute (xyz), (yzx), (zxy)
- C vCx2 Cy2 Cz2
- AB sin ?
- Directed by RH Rule.
8Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Vector Product
- A x B - B x A
- A x ( B C ) A x B A x C
- d/dt ( A x B ) d A /dt x B A x d B/dt
- i x i j x j k x k 0
- i x j - j x i k
- j x k - k x j i
- k x i - i x k j
9Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Angular Momentum
- L r x p
- The angular momentum is the vector product of
the moment arm and the linear momentum. - ? T d L/dt
- The net torque is equal to the time rate of
change in the angular momentum.
10Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Proof
- ? T r x ?F r x d p/dt
- And
- d L/dt d( r x p) /dt
- d r/dt x p r x d p/dt.
- But p m d r/dt , therefore d r/dt x p 0
- d L/dt r x d p/dt
- And thus
- ? T d L/dt.
11Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Rotating Platform Demonstration
12Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Analysis
- Why does an ice skater increase her angular
velocity without the benefit of a torque? - L r x p
- r x ( m v)
- r x ( m r x ?)
- Li mi ri 2 ?
- Lz (?i mi ri 2 ) ?
- Lz I ? ? Lz / I
- Therefore, a decrease in I ( by reducing r) will
result in an increase in ?.
13Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Torque and Angular Momentum (redux)
- L I ?
- The quantity I is a tensor .
- Changing I changes the magnitude and alignment
of L and ?.
14Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- If I is constant then for rotation about an axis
z - ? Tz d Lz /dt Iz ?
- In the absence of net external torque the angular
momentum is conserved.
15Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Spinning Bicycle Wheel
- Demonstration
16Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
17Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Fundamental Angular Momentum
- Fundamental unit of angular momentum ?
- ? 1.054 x 10 -34 kg?m/s2
- ICM? ?
- ? ? / ICM
- 1.054 x 10 -34 kg?m/s2 / (1.95 x 10 -46 kg?m)
- 5.41 x 10 11 rad/s
18Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Summary
- The total Kinetic energy of a rotating system is
the sum of the rotational energy about the Center
of Mass and the translational KE of the CM. - K ½ ICM ? 2 ½ MR 2 ? 2
- T r x F
-
19Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Summary Question
- Which will win a race down an incline?
- A ball rolling without slipping?
- A ball sliding with no friction?
- Why?
-
K ½ ICM ? 2 ½ MR 2 ? 2
20Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Summary
- Angular momentum L is the vector product of the
moment arm and the linear momentum. - L r x p
- The net externally applied torque is equal to
the time rate of change in the angular momentum. - ? Tz d Lz /dt Iz ?
21Physics 1710Chapter 12 Rolling Motion and
Angular Momentum
- Summary
- Angular momentum about an axis z is equal to
the product of the moment of inertia of the body
about that axis and the angular velocity about z. - L I ?
- Lz Iz ?
- In the absence of torques, the angular momentum
is conserved.
22Physics 1710Chapter 12 Rolling Motion and
Angular Momentum