10 o v g d Starting with sketching the coordinate vectors, ˆ e t is parallel to the velocity pointing to the right and slightly up, and ˆ e n points down and slightly to the right. Based on our kinematic expression, we can write the acceleration vector as the following. a = ˙ v ˆ e t + 1000 2 ρ ˆ e n (2.55) From the given information, we know that the acceleration vector is composed of two physical components. The drag component acts opposite the velocity vector, in the − ˆ e t direction. The gravity component acts straight down with components in both directions. a = − 1160ˆ e t + 9 . 81 (cos 10 ◦ ˆ e n − sin 10 ◦ ˆ e t ) (2.56) From these two expressions for the acceleration vector, we extract the scalar equations. ˙ v = − 1160 − 9 . 81 sin10 ◦ (2.57) 1000 2 ρ = 9 . 81 cos10 ◦ (2.58) Solving for the unknowns, ˙ v = − 1162 m/s 2 and ρ = 103 , 500 m = 103 . 5 km.
32 CHAPTER 3. PARTICLE KINETICS 3.3 Example Problems Example 3.1 Consider a fuzzy die hanging on a cord from the rearview mirror of a car. The car is decelerating at 0 . 2 g . Find the steady-state angle the cord makes with the vertical. θ i ^ j ^ 1. Kinematics - Let’s use the coordinate vectors indicated in the diagram. In the steady-state condition, the die’s acceleration is the same as the car: a = 0 . 2 g ˆ i . 2. FBD - In sketching your own FBD, the only forces acting on the die are gravity, mg , and the tension in the cord, T . The resulting force expression is Σ F = T sin θ ˆ i + ( T cos θ − mg ) ˆ j . 3. N2L - Substituting into Newton’s second law: T sin θ ˆ i + ( T cos θ − mg ) ˆ j = 0 . 2 mg ˆ i (3.5) Extracting scalar equations of motion: T sin θ = 0 . 2 mg (3.6) T cos θ − mg = 0 From these equations, solve for θ . tan θ = 0 . 2 ⇒ θ = 11 . 31 ◦ (3.7)
34 CHAPTER 3. PARTICLE KINETICS Example 3.3 The rod rotates about B in the horizontal plane with a constant angular rate ˙ θ = 0 . 05 rad/s. Collar A has a mass of 0.3 kg and slides along the rod without friction. At the current instant, the collar is at a radius of r = 0 . 15 m and is sliding along the rod with ˙ r = 0 . 2 m/s. Find the value of ¨ r and the force exerted on the collar by the bar. 0.15 m θ = 0.05 rad/s . A B 1. Kinematics - Choosing polar coordinate vectors, ˆerpoints along the rod, andˆeθpoints up and to the left.Using the given information, we can write anexpression for the acceleration vector. 2. FBD - In sketching your own FBD, the rod exerts a force N on the collar. The force is perpendicular to the rod, but we’re not sure if it’s in the positive or negative ˆ e θ direction. So for now, let’s just guess it’s in the positive ˆ e θ direction. Σ F = N ˆ e θ (3.14) Of course, gravity is also acting on the collar. But since we’re focused on the motion in the horizontal plane, we can neglect gravity. 3. N2L - Note that the collar has a mass of 0.3 kg. N ˆ e θ = 0 . 3 ( ¨ r − 3 . 75 · 10 − 4 ) ˆ e r + 0 . 02ˆ e θ (3.15) Extracting scalar equations of motion: 0 = 0 . 3 ( ¨ r − 3 . 75 · 10 − 4 ) (3.16) N = 0 . 3 · 0 . 02 From these equations, ¨ r = 3 . 75 · 10 − 4 m/s 2 and N = 0 . 006 N. We had guessed that N acts in the positive ˆ e θ direction. Since we got a positive answer for N , this guess must have been correct.
48 CHAPTER 3. PARTICLE KINETICS Example 3.11 Block A has a mass of 50 kg, and block