PEP 294: Solutions

XI. Angular Kinematics

Introductory Problems (pp. 354-356)

1. Squats: 180° <=> 95°, 10 squats; Compute the overall angular distance & displacement.

(Tip -- Divide the squat cycle into sub-intervals that show rotations in different direction: squat-down and squat-up.)

- Figure out the angular displacements for the squat-down & squat-up phases first:

Squat down phase:

q₁ = 180°, q₂ = 95°

Dq_down = q₂ - q₁ = 95 - 180 = -85°

Squat up phase:

q₁ = 95°, q₂ = 180°

Dq_up = q₂ - q₁ = 180 - 95 = 85°

- Compute the overall angular displacement and distance for a complete squat cycle:

Dq_cycle = Dq_down + Dq_up = (-85) + (85) = 0°

Angular distance per cycle = |Dq_down| + |Dq_up| = 85 + 85 = 170°

- Compute the overall angular displacement for 10 complete squat cycles:

Dq_overall = (Dq_cycle)(10 cycles) = (0)(10) = 0° = 0 rad

Angular distance = (angular distance per cycle)(10 cycles) = (170)(10) = 1700 ° = 29.67 rad

2. q₁ = 12 o'clock = 90°, q₂ = 6 o'clock = -90°, Dt = 30 s; Compute Dq.

Tip -- Note that angular position q₂ is -90° rather than 270° since the second hand shows a clockwise rotation. The angular position should be continuous as hand passes the 0° line (horizontal axis, 3 o'clock). Thus the angular position at 6 o’clock should be -90°.

- Compute the angular displacement, velocity and acceleration:

Dq = q₂ - q₁ = (-90°) - 90° = -180° = -p rad

-6 °/s =

a = 0 °/s² = 0 rad/s² (the second hand rotates with constant speed)

3. w = 2p rad/s, Dt = 20 s; Compute Dq.

Use the definition of angular velocity:

  => 

Dq = (3p)(20) = 60p rad = 30 rev (since 1 rev = 2p rad)

4. Kicking: Dt = 0.4 s, counterclockwise rotation, a = 200 °/s²; Compute w at contact.

Tip -- The initial angular velocity is 0 °/s.

Compute w_contact using the definition of angular acceleration:

; 

w_contact = (200)(0.4) = 80 °/s = 1.4 rad/s

5. w₁ = 3 rad/s, w₂ = 2.7 rad/s, Dt = 0.5 s; Compute a.

-0.6 rad/s²

10. Racket swing:

v = r^.w = (0.5)(12) = 6 rad/s

 

1. r = 1.2 + 0.76 = 1.96 m, v = 35 m/s; Compute w.

v = r^.w;  35 = (1.96)(w)

17.86 rad/s

2. r = 0.75 m, a = 20 rad/s², Dt = 1.5 s; Compute v_release.

Tip -- The initial angular velocity is 0 rad/s. Compute angular velocity at release first and then compute the release velocity.

- Angular velocity at release (w_release):

; 

w_release = (20)(1.5) = 30 rad/s

- Compute v_release:

v_release = r^.w_release = (15)(1.5) = 22.50 m/s

3. r = 46 cm = 0.46 m, w = 70 rad/s, q = 45°, d_h = 110 m; See if d_v > 0.

Tip -- This question consists of two parts: (a) to figure out the initial velocity of the projectile motion of the ball, and (b) to figure out the height of ball (d_v) when the horizontal position (d_h) of the ball is 110 m. Note that the projection height and the height of the fence are the same (1.2 m), which simplifies the situation.

- First, compute the projection velocity:

v = r^.w = (0.46)(70) = 32.2 m/s

- Resolve the velocity vector into components:

v = ( v_x, v_y ) = ( v^.cos q, v^.sin q ) = ( (32.2)(cos 45°), (32.2)(sin 45°) ) 

= ( 22.77, 22.77 ) m/s

- Compute the time when d_h = 110 m from the plate (t₁₁₀) using formula (2) for the horizontal motion,

d_h = v_x^.t;  110 = (22.77)^.t₁₁₀

- Compute the height of the ball (d_v) at t₁₁₀ using formula (2) for the vertical motion:

-4.45 m

The fact that the height of the ball at d_h = 110m is negative means the ball falls down to the ground before it reaches the fence. Thus, it can not clear the fence.

4. r = 2.5 m, w = 1.0 rad/s, v_horse = 5 m/s; Compute v_ball.

- First, compute the linear velocity of the stick head due to rotation of the stick-arm compound:

v_head = r^.w = (2.5)(1.0) = 2.5 m/s

- The velocity of the stick head due to rotation has the same direction to the velocity of horse, so simply add two velocities to compute the resultant velocity:

v_ball = v_head + v_horse = 2.5 + 5.0 = 7.5 m/s

5. v_head = 2.5 m/s, v_horse = 5.0 m/s, angle = 30°; Compute v_ball.

- Using the law of cosines:

= 7.27 m/s

- Law of sines:

; 

q = sin^-1(0.172) = 9.90°

7. w₁ = 2.5 rev/s, a = -0.2 rev/s², w₂ = 0.8 rev/s; Compute q.

- Revisiting formula (3) for the vertical motion of projectile motion:

v₂² - v₁² = 2(-g)d_v

Replace v with w, (-g) with a, and d_v with q:

w₂² - w₁² = 2aq

This is the angular version of formula (3) of constant acceleration motion. Use this equation to solve this problem.

- Compute q:

(0.8)² - (2.5)² = (2)(-0.2)^.q

14.03 rev

Note that the unit of angular displacement is ‘rev’ since the angular velocities and acceleration are given in rev/s and rev/s², respectively.