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Physics 101: Lecture 17 Rotational Dynamics

Physics 101: Lecture 17 Rotational Dynamics. Review of the last lecture on Rotational Kinematics and Rolling Motion and the Vector Nature of Angular Variables Today’s lecture will cover Textbook Sections 9.1 - 9.3. But first for something completely different ….

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Physics 101: Lecture 17 Rotational Dynamics

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  1. Physics 101: Lecture 17Rotational Dynamics • Review of the last lecture on Rotational Kinematics and Rolling Motion and the Vector Nature of Angular Variables • Today’s lecture will cover Textbook Sections 9.1 - 9.3

  2. But first for something completely different … The Quantum and the Cosmos By Rocky Kolb Today, 4:30 pm, 114 Hochstetter Dark matter ? How did our universe evolve out of a formless, shapeless quantum soup of “elementary” particles ? Dark Energy ?

  3. See text: chapter 8 And for a point at a distance R from the rotation axis: • x = Rv = R a = R Rotation Summary (with comparison to 1-D linear kinematics) Angular Linear See Table 8.1

  4. Rolling Motion • Rolling motion involves both linear and rotational motion. If there is no slipping at the point of contact with the ground, the motion of an automobile tire is an example for rolling motion. While the car is moving with a linear speed v covering a distance d, a point on the outer edge of the tire moves the same distance along the circular path s = q r = d. Thus, the linear speed v=d/t of the car is the same as the tangential speed of a point on the outer edge on the tire vT=s/t=q/t r = w r: v = vT = w r (w in rad/s)

  5. Vector Nature of Angular Variables • Like linear velocity and acceleration, also angular velocity and acceleration are vector quantities. So far we only talked about the magnitude of these vectors. But as vectors they also have a direction. Both angular velocity and acceleration point along the rotation axis. As in linear motion, if the velocity is decreasing, the acceleration points opposite to the angular velocity and if the velocity is increasing, velocity and acceleration point in the same direction.

  6. Rotational Dynamics • Remember Newtons first and second laws: An object does not change its state of motion unless a net force is applied. If a net force is applied on an object with mass m it experiences an acceleration: a = Fnet/m or Fnet=m a What is the equivalent of force and mass (inertia) in rotational motion of a rigid object ?

  7. See text: chapter 9 New Concept: Torque • Torque is the rotational analog of force: • Torque = (magnitude of force) x (lever arm) • t = F l SI unit: [N m] The lever arm is the distance between the axis of rotation and the “line of action”, i.e. the line drawn along the direction of the force. The torque is positive, if resulting into a counter clockwise rotation and negative for a clockwise rotation.

  8. CORRECT Concept Question The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case. In which of the cases is the torque on the nut the biggest? 1. Case 1 2. Case 2 3. Case 3

  9. Center of gravity pivot d W=-mg Rigid Objects in Equilibrium A system is in equilibrium if and only if: • a= 0  SFext = 0 •  = 0  S ext = 0 (about any axis) • Torque about pivot due to • gravity: • g = m g d • (gravity acts at center of gravity) This object is NOT in equilibrium.

  10. Center of Gravity • The center of gravity of a rigid body is the point at which the gravitational force is considered to act when the torque due to gravity (weight) is calculated. If an object is of symmetric shape and the weight is distributed uniformly, the center of gravity is located at the geometrical center of the object. If the object is made of a group of objects with known weights, then its center of gravity is determined by the sum of the torques around an arbitrary axis: xcg =(W1 x1 + W2 x2 +W3 x3 + …)/(W1+W2+W3+..) xcg is the point at which the weight W1+W2+W3+… acts.

  11. pivot Center of gravity pivot d Center of gravity W=-mg Torque about pivot = 0 Torque about pivot  0 Not in equilibrium Equilibrium A method to find center of gravity of an irregular object

  12. Torque • Painter on plank (uniform) which sits on two supports. • Free body diagram for plank: FA FB -Mg -mg To compute the net torque, choose a rotation axis, e.g. let’s consider the pivot is located at support A.

  13. Torque • If its just balancing on “B”, then FA = 0 • the only forces on the beam are: x FB Mg mg Using FTOT = 0: FB = Mg + mg This does not tell us x

  14. Torque • Find net torque around pivot B: (or any other place) FB d1 d2 Mg mg t(FB ) = 0 since lever arm is 0 t(Mg) = Mgd1 Total torque = 0 = Mgd1-mgd2 t(mg) = -mgd2 So d2 = Md1 /mand you can use d1 to find x

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