Unformatted text preview: Even if an object is not in actual orbit (where it never moves any closer to the planet), any circular path, even for a small part of a circle, indicates the net force on the object is essentially centripetal during that time—and gravity can be the sole supplier of that force. A roller-coaster features a full “loop-the-loop” (where the train and all its passengers are momentarily upside-down at the top of the vertical circle). It is designed so that passengers don’t actually need their safety harnesses while upside-down; their motion would still (barely) keep them seated at that point. If such a train has a speed of 12 m/s at the top of the loop, what is the radius of the loop? 1. 2. 3. 4. 5.
11/11/09 Approximately 14.6 m. Approximately 12.2 m. Approximately 8.17 m. Approximately 6.81 m. None of the above.
Oregon State University PH 201, Lecture #20 The center of gravity of a rigid body is the point at which the force of gravity appears to act on the body—as if all of the body’s mass were concentrated at that point. The key to balancing the meter stick is restoring the stick’s center of gravity to its original position at the fulcrum (where it was before weights were added). When the 5-N weight was added, that center moved to the right. What calculation were we doing to ﬁgure out what weight to place at the 10-cm position to restore the center of mass to the 50-cm position? 11/11/09 Oregon State University PH 201, Lecture #20 What is a torque? It is a “twisting” force that causes angular acceleration in a rigid body—usually caused by a linear force acting at some point on the body, causing it to rotate about some axis. It is computed as = F·l, where l is the perpendicular distance from the body’s axis of rotation to the force’s line of action. An easy way to compute this is = (Fsin )d where d is the line segment from the axis of rotation to the point where F is acting on the body, and is the angle between F and d. Torque has units of force·distance (N·m). Torque is a vector. A torque is considered a positive value when its twisting effect is in the counter-clockwise direction; negative if clockwise.
11/11/09 Oregon State University PH 201, Lecture #20 Notice that in balancing the torques acting on the meter stick, we prevented the stick from accelerating—in the angular sense. Indeed, Newton’s First Law must include rotational motion, too: A rigid body is in total mechanical equilibrium only when it is not accelerating along any translational axis; and when it is not accelerating around any rotational axis. In other words, it’s in total mechanical equilibrium when: F=0 and =0 11/11/09 Oregon State University PH 201, Lecture #20 What about Newton's 2nd Law? Is there a rotational version of this, too? Yes. As we know, F = ma. What would be the rotational analogy to that? This: =I What net torque, , exerted about the axis of rotation, is needed in order to cause an angular acceleration, , of the object about that axis? It’s the moment of inertia (about that axis) for that object, multiplied by the desired angular acceleration. 11/11/09 Oregon State University PH 201, Lecture #20 Here are some of the direct analogies between (linear) translational and rotational motion: Quantity or Principle Displacement Velocity Acceleration Inertia (resistance to acceleration) Momentum Cause of acceleration Newton’s 2nd Law
11/11/09 Linear x v a mass (m) P = mv force (F) F = ma Rotation moment of inertia (I) L=I torque ( ) =I Oregon State University PH 201, Lecture #20 ...
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