Review of physics concepts for Exam 3. April, 2019

Transcription

1 Review of physics concepts for Exam 3 April, 2019

2 Reminders: 1. The vector sum of all forces = (the total inertial mass ) *a 2. Gravity F = mg; E=mgh 3. Friction along a surface Ff = (friction coefficient) * (force normal to the surface) = µ N New: 1. Momentum 2. Impact 3. Circular motion 4. Circular energy

3 1. Conservation of momentum in collision Momentum= p = m v Since v is a vector, p is too p(before) = p(after) p1+p2 = p1 + p2

4 1. crash The momenta of cars A and B are 3 and -4 in units of kg m/s before they collide. What is the total momentum afterwards in kg m/s? a. 1 b. 3 c. 4 d. 7 e. -1 Momentum before = 3-4= -1 After = same

5 2. Heavy and light cars collide A light car B going East collides with a stationary heavy car A. They don t stick to each other. What is the most likely direction in which B the light car will go after the collision? a. N b. S c. E d. W e. Can t tell Car A is heavy. It s like hitting a wall. Car B goes back => W A

6 3. Impact The size of an impact depends on the change of momentum, p A, of the car, A (or other thing). The force in an impact is bigger if the impact is sharp (short). Impact = force times time: F t The impact is equal to the momentum before momentum after F t = p A p A

7 3. Impact A ball hits a wall elastically and straight on. The impact takes 0.1 s. The average force in the collision is F = 100 N. What is the ball s momentum after the collision? a. 5 b. 10 c. 50 d. 100 e F t = p (-p) => 2p = Ft => p = 100 * 0.1 / 2 = 5

8 4. Circular constant linear speed Speed = distance / time = circumference / period for one revolution: v =2 π R / T For n revolutions: v = n 2 π R / t

9 4. Circular constant speed You swing a ball around in a circle with a circumference of 6 m. You time it for 10 swings and find it takes 5 s. What is the average speed in m/s? a. 6 b. 10 c. 5 d. 3.1 e. 12 v = n 2 π R / t = 10 * 6 / 5 =12

10 5. Circular acceleration 3 kinds of acceleration: 1. Linear = change in speed / t = a **direction: straight ahead 2. Angular = change in angle /t = α **direction: right hand rule 3. Circular = change in direction / t = a = v 2 /R ** direction : in Constant v a = v 2 / R v a R

11 5. Circular acceleration You swing a ball around in a circle at constant speed of 2 m/s. You measure the radius to be ½ m. What is the circular acceleration of the ball in m / s 2? a. 2 b. 4 c. 6 d. 8 e. 10 a= v 2 / R = 2*2/ 0.5 = 8

12 6. Force to keep a ball moving in a circle At constant v, the force is the mass of the ball times the acceleration. a and F both point in. F = m a = m v 2 / R a F

13 6. Force to hold in a ball moving in a circle You swing a ball of mass 1/2 kg around in a circle using a force on a wire attached to the ball to hold it at a constant radius 0.5 m. The ball has a constant speed of 2 m/s. What is the force, in N? a. 2 b. 4 c. 6 d. 8 e. 10 F= m v 2 /R = 0.5 * 2 * 2 / 0.5 = 4

14 7. Circular motion with any speed Once around: Linear Speed =x/t Angular distance: θ=2π Angular speed = θ/t v =2πR/T ω =2π/T Any n revolutions: Linear Speed v = n 2πR/t in m / s Angular distance: θ=n 2π Angular speed ω = n 2π/t in Radians / s

15 7. Angular distance and time You are swinging a ball in a circle of radius R = ½ m for a time of 31.4 s. You count that it goes around for 10 times. What is the average angular speed in Radians / s? a. 10 b. 5 c. 2 d. 1 e. 0.5 ω = n 2 π / t = 10 * 2 * 3.14 / 31.4 = 2

16 8. Angular speed and acceleration and both Linear: v = x / t (v + v 0 )/2 = v avg = (x x 0 ) / (t - t 0 ) Angular: ω = θ / t (ω + ω 0 )/2 = ω avg = θ / t a = (v v 0 )/t α= (ω ω 0 )/t x = v 0 t + ½ a t 2 θ = ω 0 t + ½ α t 2 2 a x = ( v 2 v 02 ) 2 α θ = (ω 2 ω 02 )

17 8. Angular speed A spinner slows to a stop with a constant acceleration in a time of 8 s and an average speed of 2 Radians/s. Through what angle does it turn, in Radians? a. 8 b. 4 c. 16 d. 2 e. 32 ω avg = θ /t => θ = ω avg t = 16 Radians (about 2.5 turns)

18 9. Reminder about Angular distance, speed and acceleration Linear: v = x / t (v + v 0 )/2 = v avg = (x x 0 ) / (t - t 0 ) Angular: ω = θ / t (ω + ω 0 )/2 = ω avg = θ / t a = (v v 0 )/t α= (ω ω 0 )/t x = v 0 t + ½ a t 2 θ = ω 0 t + ½ α t 2 2 a x = ( v 2 v 02 ) 2 α θ = (ω 2 ω 02 )

19 9. Angular acceleration A spinner slows from an initial angular speed of 4 Radians/s to a stop in 2 seconds. What is its angular acceleration in Radians/s 2? a. 2 b. 4 c. 8 d. -2 e. -4 α = ( ω ω 0 ) /t = (0 4)/2 = -2

20 10. Reminder about Angular distance, speed and acceleration Linear: v = x / t (v + v 0 )/2 = v avg = (x x 0 ) / (t - t 0 ) Angular: ω = θ / t (ω + ω 0 )/2 = ω avg = θ / t a = (v v 0 )/t α= (ω ω 0 )/t x = v 0 t + ½ a t 2 θ = ω 0 t + ½ α t 2 2 a x = ( v 2 v 02 ) 2 α θ = (ω 2 ω 02 )

21 10. Angle and circular acceleration A spinner slows from an initial angular speed of 4 Radians/s. It slows for 8 s with a constant acceleration of ½ Radians/s 2. Through what angle does it turn, in Radians? a. 16 b. -16 c. 32 d. 64 e. -64 θ = ω 0 t + ½ α t 2 = 4 * 8 + ½ (- ½ ) 8 2 = = 16

22 11. Reminder about Angular distance, speed and acceleration Linear: v = x / t (v + v 0 )/2 = v avg = (x x 0 ) / (t - t 0 ) Angular: ω = θ / t (ω + ω 0 )/2 = ω avg = θ / t a = (v v 0 )/t α= (ω ω 0 )/t x = v 0 t + ½ a t 2 θ = ω 0 t + ½ α t 2 2 a x = ( v 2 v 02 ) 2 α θ = (ω 2 ω 02 )

23 11. Rotation with no time A spinner slows to a stop from an initial angular speed of 3 Radians/s. It slows with an acceleration of ¼ Radians/s 2. Through what angle does it turn, in Radians? a. 36 b. -36 c. 18 d. -18 e α θ = (ω 2 ω 02 ) => θ = (ω 2 ω 02 ) /2 α = ( 0 3*3)/2 (- ¼ ) =18

24 12. Moment of Inertia: I I plays the role of the inertial mass in rotation. The kinetic energy = ½ I (ω 2 ω 02 ) I is in general of the form: c m R 2 For any small object swinging around on a string: I= 1 m R 2 For a ball rotating about its center: I = 2/5 m r 2 For a ball swinging around on a string: I = 1 m R 2 + 2/5 m r 2 For a ball rolling: : I = 1 m r 2 + 2/5 m r 2 All the values of c that you need are in the equation sheet.

25 12. Moment of inertia A bowling ball has mass 5 kg and radius 0.12 m. What is its Moment of inertia it you spin it about its axis? a. 0.6 b c. 5 d e I = c m r 2 = 2/ = 2 144/10000 = 0.03

26 13. Rotational energy v = x / t (v + v 0 ) = v avg = (x x 0 ) / (t - t 0 ) ω= θ / t a = (v v 0 )/t α= (ω ω 0 )/t x = v 0 t + ½ a t 2 θ = ω 0 t + ½ α t 2 2 a x = ( v 2 v 02 ) => W = ma d = ½ m ( v 2 v 02 ) = K 2 α θ = (ω 2 ω 02 ) => W = I α θ = ½ I (ω 2 ω 02 ) = K

27 13. Rotational kinetic energy A student spins a bowling ball, with moment of inertia I = 0.03 kg m 2, about its axis. It starts from rest and reaches a speed of 10 radians/s. What is its rotational energy change in J? a. 1.5 b. 3 c d e. 0.3 K= ½ I ω 2 = ½ = 1.5