W13D1-1 Reading Quiz and Concept Questions

Transcription

1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Fall Term 2009 W13D1-1 Reading Quiz and Concept Questions A person spins a tennis ball on a string in a horizontal circle (so that the axis of rotation is vertical). At the point indicated below, the ball is given a sharp blow vertically downward. In which direction does the angular momentum about the center of the circle change. 1. +x direction 2. -x direction 3. +y direction 4. -y direction 5. It stays the same (but the magnitude of the angular momentum about the center of the circle changes). 6. The ball starts wobbling in all directions. Answer 1. According to the right-hand rule, the torque exerted by the force is in the forward direction (+x), and so the change in angular momentum must also be in this direction.

2 Note: As it receives the blow, the ball accelerates downward and its velocity gains a downward component. Still constrained by the string to go around in a circle, the ball moves in a new circular path the plane which is determined by the instantaneous velocity v the ball has immediately after the blow. The direction of this velocity v' and the direction along the string define a plane that is tilted downward in the forward direction (+x direction). The angular momentum therefore also tilts forward in the +x direction.

3 Concept Question: Chasles Theorem The displacement of any rigid body can be described by the displacement of the center of mass and a rotation about some axis passing through the center of mass 1. This is not true. 2. This is always true. 3. This is only true for small (differential displacements) 4. It is only true if the rotation is about a symmetry axis. Answer 2. The key condition is that the body is a rigid body. For a more detailed discussion and proof of this theorem see

4 Concept Question All points on a body rotate at the same 1. For all bodies. 2. For all bodies only when ω r goes through the CM. 3. For all rigid bodies. 4. For all rigid bodies only when ω r goes through the CM. Answer 3. The key point again is that the body is rigid. It doesn t matter where the axis of rotation passed through the body.

5 Concept Question A non-symmetric body rotates with constant angular velocity about its CM. Relative to the CM 1. L r is constant. 2. L r r is constant but L / r r 3. L / L is constant but L r 4. L r lies along r ω. r L is not. is not. Answer 2. The angular momentum about the CM does not point along the axis of rotation and it precesses as the body rotates so the direction of the angular momentum about the CM, r r L / L, is not constant but the magnitude of L r is constant.

6 Concept Question A rigid body with rotational symmetry body rotates at a constant angular velocity ω r about it symmetry (z) axis. In this case 1. L r is constant. 2. L r r is constant but L / r r 3. L / L is constant but L r 4. L r lies along r ω. 5. Two of the above are true. r L is not. is not. Answer r 5. The magnitude and direction of ω so both 1 and 4 are true. L r is constant and it points in the same direction as

7 Concept Question Consider a massless rod of length 2l with two point-like objects of mass m at each end, rotating about the vertical z-axis with angular velocity ω r. There is a sleeve on the axis of rotation. At the moment shown in the figure, two forces are acting on the sleeve. The torque about the center of mass due to the forces points 1. along the z-axis. 2. along the line formed by the rod. 3. in the plane of the figure, perpendicular to the line formed by the rod. 4. into the plane of the figure. 5. out of the plane of the figure. Answer 4: The total torque is given by r r r r r τ = F + F,1 1,2 2 Each term in the above expression points into the plane of the figure.

8 Concept Question. Rotating Disk Pendulum A rotating disk pendulum consists of a uniform spinning diskon a shaft attached to a pivoted rod. Assume that the rod and shaft are massless. The uniform disk spins on a frictionless bearing. The pendulum is dropped from rest in a horizontal position. What happens to the magnitude of ω of the spinning disk as the pendulum drops? 1) ω increases 2) ω decreases 3) ω remains the same Answer 3. Since the disk spins on a frictionless bearing there is no torque on the disk that will cause ω to change magnitude.

9 Concept Question: Rotating Disk Pendulum A rotating disk pendulum consists of a uniform spinning disk on a shaft attached to a pivoted rod. Assume that the rod and shaft are massless. The uniform disk spins on a frictionless bearing. The pendulum is dropped from rest in a horizontal position. At the bottom of the swing, when the rod is vertical 1) Ω increases 2) Ω decreases 3) Ω remains the same Answer 3. If the only forces acting on the disk pendulum are the two shown in the figure and the gravitational force on the center of mass so at the bottom of the swing there is no torque about eh origin due to gravitation and the torque about the origin due to F b and F c is along the y-axis (it will point in the negative y-direction). Since there is no torque along the x-axis, Ω remains the same.