Concept Question: Normal Force


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1 Concept Question: Normal Force Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is 1. larger than 2. identical to 3. smaller than the downward force of gravity on the person.
2 Concept Question: Normal Force Answer Answer 1. The normal force on the person is greater than the gravitational force on the person because the normal force must also accelerate the person, as well as oppose the gravitational force. Thus N mg = ma implies that N > mg. So the magnitude of upward normal force is larger than the magnitude of the downward gravitational force.
3 Concept Q.: Constraints and Pulleys A massless rope is attached to the ceiling at one end and held stationary with tension T at the other end. Through a system of frictionless pulleys it supports a body of mass m. What is the value of m? T / g 4T sinθ / g (2 + 2)T / g (2 + 2)T sinθ / g 5T / g
4 Con. Q. Ans.: Constraints and Pulleys Answer 3. The object is static, so the vertical sum of the forces acting on the object is zero 2T + 2T sin 45 mg = 0 Thus the value of the mass is m = (2 + 2)T / g
5 Concept Question: Angular Speed Object A sits at the outer edge (rim) of a merrygoround, and object B sits halfway between the rim and the axis of rotation. The merrygoround makes a complete revolution once every thirty seconds. The magnitude of the angular velocity of Object B is 1. half the magnitude of the angular velocity of Object A. 2. the same as the magnitude of the angular velocity of Object A. 3. twice the the magnitude of the angular velocity of Object A. 4. impossible to determine.
6 Concept Question Ans.: Angular Speed Object A sits at the outer edge (rim) of a merrygoround, and object B sits halfway between the rim and the axis of rotation. The merrygoround makes a complete revolution once every thirty seconds. The magnitude of the angular velocity of Object B is Answer: 2. All points in a rigid body rotate with the same angular velocity.
7 Concept Question: Circular Motion An object moves counterclockwise along the circular path shown below. As it moves along the path its acceleration vector continuously points toward point S. The object 1. speeds up at P, Q, and R. 2. slows down at P, Q, and R. 3. speeds up at P and slows down at R. 4. slows down at P and speeds up at R. 5. speeds up at Q. 6. slows down at Q. 7. No object can execute such a motion.
8 Concept Question Circular Motion: Answer Answer 3. At the point P the acceleration has a positive tangential component so it is speeding up. At the point S the acceleration has a zero tangential component so it is moving at a constant speed. At the Point R the acceleration has a negative tangential component so it is slowing down.
9 Concept Question: Circular Motion and Force A pendulum bob swings down and is moving fast at the lowest point in its swing. T is the tension in the string, W is the gravitational force exerted on the pendulum bob. Which freebody diagram below best represents the forces exerted on the pendulum bob at the lowest point? The lengths of the arrows represent the relative magnitude of the forces.
10 Concept Question Circular Motion and Force: Answer Answer d. The bob is undergoing circular motion. It is accelerating towards the center. Newton s Second Law gives T W = mrω 2 hence T = W + mrω 2 so T > W.
11 Concept Q.: Pushing a Baseball Bat A baseball bat is pushed with a force F. You may assume that the push is instantaneous. Which of the following locations will the force produce an acceleration of the center of mass with the largest magnitude? 1. Pushing at Position Pushing at Position 2 (center of mass). 3. Pushing at Position Pushing at 1,2,and 3 all produce the same magnitude of acceleration of the center of mass/
12 Con. Q. Ans. : Pushing a Baseball Bat Answer 4. The external force is equal to the total mass times the instantaneous acceleration of the centerofmass. It doesn t matter where the external force acts with regards to the centerofmass acceleration.
13 Concept Question: Losing Mass But Not Momentum Consider an ice skater gliding on ice holding a bag of sand that is leaking straight down with respect to the moving skater. As a result of the leaking sand, the speed of the skater 1. increases. 2. does not change. 3. decreases. 4. not sure.
14 Con. Q. Ans.: Losing Mass But Not Momentum Answer 2. The sand leaves the bag with the same horizontal speed as the skater. The momentum of the system of the skater and sand does not change and so the speed of the skater does not change.
15 Concept Question: Pushing Carts Consider two carts, of masses m and 2m, at rest on an air track. If you push one cart for 3 seconds and then the other for the same length of time, exerting equal force on each, the kinetic energy of the light cart is 1) larger than 2) equal to 3) smaller than the kinetic energy of the heavy car.
16 Concept Question Ans. : Pushing Carts Answer 1. The kinetic energy of an object can be written as K = 1 2 mv2 = p2 2m Because the impulse is the same for the two carts, the change in momentum is the same. Both start from rest so they both have the same final momentum. Since the mass of the lighter cart is smaller than the mass of the heavier cart, the kinetic energy of the light cart is larger than the kinetic energy of the heavy cart.
17 Concept Question: Potential Energy and Inverse Square Gravity A comet is speeding along a hyperbolic orbit toward the Sun. While the comet is moving away from the Sun, the change in potential energy of the Suncomet system is: (1) positive (2) zero (3) negative
18 Concept Question Answer: Potential Energy and Inverse Square Gravity Answer: 1. The displacement of the comet has a component in the opposite direction as the force on the comet so the work done is negative. (The comet's acceleration is always toward the Sun; when the comet moves away from the Sun, the work is negative.) Therefore the change in potential energy of the Suncomet system is positive.
19 Concept Question: Recoil Suppose you are on a cart, initially at rest on a track with very little friction. You throw balls at a partition that is rigidly mounted on the cart. After the balls bounce straight back as shown in the figure, is the cart 1. moving to the right? 2. moving to the left? 3. at rest.
20 Concept Question: Recoil Answer: 2. Because there are no horizontal external forces acting on the system, the momentum of the cart, person and balls must be constant. All the balls bounce back to the right, then in order to keep the momentum constant, the cart must move forward.
21 Concept Q.: Moment of Inertia Same Masses All of the objects below have the same mass. Which of the objects" has the largest moment of inertia about the axis shown?" (1) Hollow Cylinder (2) Solid Cylinder (3)Thinwalled Hollow Cylinder
22 Concept Q. Ans.: Moment of Inertia Same Masses All of the objects below have the same mass. Which of the objects" has the largest moment of inertia about the axis shown?" Answer 3. The mass distribution for the thinhollow walled cylinder is furthest from the axis, hence it s moment of inertia is largest.
23 Concept Question: Kinetic Energy A disk with mass M and radius R is spinning with angular speed ω about an axis that passes through the rim of the disk perpendicular to its plane. Moment of inertia about cm is (1/2)M R 2. Its total kinetic energy is: 1. (1/4)M R 2 ω 2 2. (1/2)M R 2 ω 2 3. (3/4)M R 2 ω 2 4. (1/4)M Rω 2 5. (1/2)M Rω 2 6. (1/4)M Rω
24 Concept Q. Ans.: Kinetic Energy Answer 3. The parallel axis theorem states the moment of inertia about an axis passing perpendicular to the plane of the disc and passing through a point on the edge of the disc is equal to I edge = I cm + mr 2 The moment of inertia about an axis passing perpendicular to the plane of the disc and passing through the center of mass of the disc is equal to I cm = (1/ 2)mR 2 Therefore I edge = (3 / 2)mR 2 The kinetic energy is then K = (1/ 2)I edge ω 2 = (3 / 4)mR 2 ω 2
25 Concept Q. Mag. of Angular Momentum In the above situation where a particle is moving in the xy plane with a constant velocity, the magnitude of the angular momentum about the point S (the origin) 1) decreases then increases 2) increases then decreases 3) is constant 4) is zero because this is not circular motion
26 Concept Q. Ans.: Mag. of Ang. Mom. Solution 3. As the particle moves in the positive x direction, the perpendicular distance from the origin to the line of motion does not change and so the magnitude of the angular momentum about the origin is constant.
27 Concept Q.: Angular Momentum of Disk A disk with mass M and radius R is spinning with angular speed ω about an axis that passes through the rim of the disk perpendicular to its plane. The magnitude of its angular momentum is: M R2 ω M R2 ω M R2 ω M R2 ω M R2 ω M R2 ω
28 Concept Q. Ans.: Ang. Mom. of Disk Answer 6. The moment of inertia of the disk about an axis that passes through the rim of the disk perpendicular to its plane is I = I cm + MR 2 = (3/2)MR 2. So the magnitude of its angular momentum is L = (3/2)MR 2 ω.
29 Concept Q.: Change in Angular Mom. A person spins a tennis ball on a string in a horizontal circle with vvelocity (so that the axis of rotation is vertical). At the point indicated below, the ball is given a sharp Fblow (force ) in the forward direction. This causes a change in angular ΔL momentum in the ˆr 1. direction ˆθ 2. direction ˆk 3. direction
30 Concept Q. Ans.: Change in Ang. Mom. ˆr Answer 3. The torque about the center of the circle points in the positive ˆk direction. The change in the angular momentum about the center of the circle is proportional to the angular impulse about the center of the circle which is in the direction of the torque
31 Concept Question: Rolling Without Slipping If a wheel of radius R rolls without slipping through an angle θ, what is the relationship between the distance the wheel rolls, x, and the product Rθ? 1. x > Rθ. 2. x = Rθ. 3. x < Rθ.
32 Concept Q. Ans. : Rolling Without Slipping Answer 2. Rolling without slipping condition, x = Rθ.
33 Concept Question: Pulling a YoYo 1
34 Concept Q. Ans.: Pulling a YoYo 1 Answer 1. For forces below a fixed maximum value, the torque about the contact point between the ground and yoyo is only due to the force F and produces an angular acceleration directed into the plane of the figure. Hence the cylinder rolls to the right, in the direction of F, winding up the string.the torque about the center of mass due to the force of friction is larger in magnitude than the torque due to the pulling force.
35 Concept Question: Cylinder Rolling Down Inclined Plane A cylinder is rolling without slipping down an inclined plane. The friction at the contact point P is 1. Static and points up the inclined plane. 2. Static and points down the inclined plane. 3. Kinetic and points up the inclined plane. 4. Kinetic and points down the inclined plane. 5. Zero because it is rolling without slipping. 35
36 Concept Q. Ans.: Cylinder Rolling Down Inclined Plane Answer 1. The friction at the contact point P is static and points up the inclined plane. This friction produces a torque about the center of mass that points into the plane of the figure. This torque produces an angular acceleration into the plane, increasing the angular speed as the cylinder rolls down. 36
37 Concept Question: Angular Collisions A long narrow uniform stick lies motionless on ice (assume the ice provides a frictionless surface). The center of mass of the stick is the same as the geometric center (at the midpoint of the stick). A puck (with putty on one side) slides without spinning on the ice toward the stick, hits one end of the stick, and attaches to it. Which quantities are constant? 1. Angular momentum of puck about center of mass of stick. 2. Momentum of stick and ball. 3. Angular momentum of stick and ball about any point. 4. Mechanical energy of stick and ball. 5. None of the above Three of the above Two of the above 14.
38 Concept Q. Ans.: Angular Collisions Answer: 7 (2) and (3) are correct. There are no external forces acting on this system so the momentum of the center of mass is constant (1). There are no external torques acting on the system so the angular momentum of the system about any point is constant (3). However there is a collision force acting on the puck, so the torque about the center of the mass of the stick on the puck is nonzero, hence the angular momentum of puck about center of mass of stick is not constant. The mechanical energy is not constant because the collision between the puck and stick is inelastic.
39 Concept Question: SHM Velocity A block of mass m is attached to a spring with spring constant k is free to slide along a horizontal frictionless surface. At t = 0 the blockspring system is stretched an amount x 0 > 0 from the equilibrium position and is released from rest. What is the x component of the velocity of the block when it first comes back to the equilibrium? 4 1. v x = x 0 2. T v x = x 0 4 T 3. v x = k 4. m x 0 v x = k m x 0
40 CQ Answer: SHM Velocity Answer 3. The particle starts with potential energy. When it first returns to equilibrium it now has only kinetic energy. Since the energy of the blockspring system is constant: (1/ 2)mv x 2 = (1 / 2)kx 0 2 Take positive root because object is moving in negative x direction when it first comes back to equilibrium, and x o > 0, we require that v x,o < 0, therefore v x = k m x 0
41 Concept Question: Energy Diagram 3 A particle with total mechanical energy E has position x > 0 at t = 0 1) escapes to infinity 2) approximates simple harmonic motion 3) oscillates around a 4) oscillates around b 5) periodically revisits a and b
42 Concept Q. Ans.: Energy Diagram 3 Solution 3. The range of motion for the particle is limited to the regions in which the kinetic energy is either zero or positive, so the particle is confined to move around the region surrounding a. The motion will be periodic but not simple harmonic motion because the potential energy function is not a quadratic function and only for quadratic potential energy functions will the motion be simple harmonic. Hence the particle oscillates around a. 42
43 Concept Question: Rotating Vector A vector A(t) of fixed length A is rotating about the zaxis at an angular speed ω. At t = 0 it is pointing in the positive y direction. da(t) / dt is given by da(t) / dt = ω Asin(ωt)î ω Acos(ωt)ĵ da(t) / dt = ω Asin(ωt)î + ω Acos(ωt)ĵ da(t) / dt = ω Asin(ωt)î ω Acos(ωt)ĵ da(t) / dt = ω Asin(ωt)î ω Acos(ωt)ĵ da(t) / dt = ω Acos(ωt)î + ω Asin(ωt)ĵ da(t) / dt = ω Acos(ωt)î + ω Asin(ωt)ĵ da(t) / dt = ω Acos(ωt)î ω Asin(ωt)ĵ da(t) / dt = ω Acos(ωt)î ω Asin(ωt)ĵ
44 Concept Q. Ans. : Time Derivative of Rotating Vector Answer 8: At time t A(t) = Asin(ωt)î + Acos(ωt)ĵ Therefore da / dt = ω Ascos(ωt)î ω Asin(ωt)ĵ and is perpendicular to A(t)
45 Concept Question: Rotating Rod Consider a massless rod of length I with two pointlike objects of mass m at each end, rotating about the vertical zaxis with angular speed ω. There is a sleeve on the axis of rotation. At the moment shown in the figure, two forces are acting on the sleeve. The direction of the change of the angular momentum about the center of mass points 1. along the zaxis. ω 2. along the line formed by the rod. 3. into the plane of the figure. 4. out of the plane of the figure.
46 Concept Q. Answer: Rotating Rod Answer 3. The torque about the center of mass points into the plane of the figure. Therefore the direction of the change of the angular momentum about the center of mass points into the plane of the figure.
47 Concept Question: Stabilizing a Turning Car When making a turn every car has a tendency to roll over because its center of mass is above the plane where the wheels contact the road. Imagine a race car going counter clockwise on a circular track. It could mitigate this effect by mounting a gyroscope on the car. To be effective the angular velocity vector of the gyro should point 1) ahead 2) behind 3) to the left 4) to the right 5) up 6) down
48 Concept Q. Ans.: Stabilizing a Turning Car Answer 4. Note that the same orientation of the gyro will stabilize the car when it turns to the right as well. The torque applied to the car by the gyro will change direction, pointing toward the rear of the car instead of the front. But now the result of the curved path is tending to roll the car to the left instead of to the right. The gyro still acts to prevent a rollover.
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