TOP VIEW. FBD s TOP VIEW. Examination No. 2 PROBLEM NO. 1. Given:

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PROBLEM NO. 1

Given: A vehicle having a mass of 500 kg is traveling on a banked track on a path with a constant radius of R = 1000 meters. At the instant showing, the vehicle is traveling with a speed of v = 50 m/sec with this speed decreasing at a rate of 4 m/sec2 (due to braking).

Find: For this instant, determine the magnitude of the TOTAL friction force acting on the vehicle by the roadway. (Here the total friction force is that due to both braking and turning.) Use the figures provided below for two views of the FBD for your analysis.

P θ = 36.87° R k en O R O en et P v

TOP VIEW

P k en

FBD’s

en et P

TOP VIEW

fturn N mg fbrake

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FBD: previous page Newton:

(1)

!

F_t = " f_brake = ma_t

(2)

!

F_n = " f_turncos#+ N sin# = ma_n

(3)

!

F_z = f_turnsin# + N cos# " mg = ma_z $

fturnsin#+ N cos# = m g + a

(

z

)

Multiply (2) by cosθ, (3) by sinθ and subtract:

f_turn

(

cos2! + sin2!

)

= "ma_ncos! + m g + a

(

_z

)

sin! # (4) f_turn = m "a$% _ncos! + g + a

(

_z

)

sin!&'

Kinematics: (5) a_t = dv dt = !4 m / sec 2 (6) a_n = v 2 R (7) az = 0 Solve: Combining (1), (4)-(7): f_brake = !ma_t = ! 500

( )

!4 = 2000 N f_turn = m !v 2 R cos" + gsin" # $ % % & ' ( (= 500

( )

! 502 1000

( )

0.8 + 9.806 0.6

( )

# $ % % & ' ( (= 1941 N Therefore,

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PROBLEM NO. 2

Given: Particles A and B (having masses of m_A = m_B = 10 kg) are interconnected by the cable-pulley system shown in the figure. Both particles are constrained to vertical motion with particle A able to slide on a smooth vertical rod. The system is released at s_A = 0 with A traveling downward with a speed of 5 m/sec. Assume the pulleys to be small, massless and frictionless.

Find: You are asked here to find the speed of particle A when A has reached the position of s_A = 2 m. In your solution clearly indicate the following of the four-step solution method:

1. FBD: Complete the free body diagram of the system of A, B and the cable shown below right. Identify any nonconservative forces that do work on the system.

2. Work-energy equation: Clearly indicate the gravitational datum line(s) used.

3. Kinematics: Here you need to relate ∆s_B to ∆s_A as well as v_B to v_A for the two positions.

4. Solve

PLEASE START YOUR WORK ON THE NEXT PAGE.

B A O Oy Ox mBg mAg N B A smooth rod g sA sB 1.5 m O datum for B datum for A

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FBD: shown on previous page. All forces doing work are conservative and will be included in the potential energy.

Work-energy: T₁= 1 2mvA1 2 ₊1 2mvB1 2 V₁= 0 T₁= 1 2mvA2 2 ₊1 2mvB2 2 V₂ = !mgs_A + mgh_B U_1"2( )nc = 0 Therefore, T₁+ V₁+ U_1!2( )nc = T₂+ V₂ " v_A12 + v_B12 = v_A22 + v_B22 + 2g #s

(

_A + h_B

)

Kinematics 2s_B + s_A2 + 1.52 = const. ! 2!s_B + sA!sA s_A2 + 1.52 = 0 ! v_B = 1 2 s_A s_A2 + 1.52 v_A Also, !s_B = 1 2 1.5 2_{+ s} A 2 _{" 1.5} # $% &'(= hB At position 1: v_B1= 1 2 0 1.5 vA1= 0 At position 2: v_B2 = 1 2 2 2.5vA2 = 0.4 vA2 and hB = 1 2

(

2.5! 1.5

)

= 0.5 m Solve: v_A2 = vA1 2 _{+ 2g s} A ! hB

(

)

1+ 0.42 = 5

( )

2_{+ 2}

( )

9.806

(

)

(

2! 0.5

)

1+ 0.16 = 6.85 m / sec

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PROBLEM NO. 3

Part (a) – 6 points

Particle B is attached to rigid bar BO with bar BO pinned to ground at O. Particle A strikes the stationary particle B with a speed of v_A1 in the direction shown. The coefficient of restitution for this impact is e < 1. Consider all surfaces to be smooth and all motion to be in a HORIZONTAL plane. Consider the following 12 statements about momentum and energy during impact and indicate if each statement is TRUE or FALSE.

HINT: Complete the FBD of each of the three systems shown below prior to responding.

For System A+B

linear momentum in the n-direction is conserved: TRUE or FALSE linear momentum in the t-direction is conserved: TRUE or FALSE angular momentum about point O is conserved: TRUE or FALSE mechanical energy is conserved: TRUE or FALSE

For System A

For System B

A B v_A1 36.87° O RIGID bar vB1 = 0 B n t B n t A n t A System A+B System A System B

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PROBLEM NO. 3

Part (b) – 6 points

An arm rotates about the fixed Z-axis with a rate of Ω. A circular disk rotates about its own axis with a constant rate of p = 4 rad/sec relative to the arm. Let XYZ represent a set of fixed coordinate axes, and xyz be a set of coordinate axis attached to the disk. At the instant shown, the xyz and XYZ axes are aligned. Also at this position, Ω = 3 rad/sec with Ω increasing at a rate of !! = 2 rad / sec2_{. For this position, determine the} angular velocity and angular acceleration of the disk. Express your answer as vectors in terms of either their xyz or XYZ

components. ! = "K + p j = 3k + 4 j

(

)

rad / sec ! = !"K + " !K + !p j + pd j dt = !"K + "0 + 0

( )

j+ p

(

# $ j

)

= 2 k + 4

( )

(

3k+ 4 j

)

$ j

)

= %12i + 2 k

(

)

rad / sec2

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PROBLEM NO. 3

Part (c) – 4 points

Particle P (having a mass of m) is able to slide on a smooth HORIZONTAL surface. An extensible cord (having a stiffness of 50 N/m and unstretched length of 2 meters) is attached between P and a fixed point O in the plane of motion for P. At position 1, P is released with a velocity as shown below with R₁ = 2 meters. Assuming that the cord remains taut for all time, find the angular speed ω of the cord about point O when P is at position 2 where R₂ = 4 meters.

HINT: consider the angular momentum of P as it moves about the fixed point O.

M_O

!

= O " H_O2 = H_O1 " m r_{P /O}! v_P2 = m r_{P /O}! v_P1 R₂e_R

(

)

! !R

(

2eR + R2"2e#

)

= R

(

1eR

)

! v

(

P1cos#eR + vP1sin#e#

)

R₂2"₂k= R₁v_P1sin#k $ !2 = R₁v_P1sin" R₂2 = 2

( )

20

( )

0.6 4

( )

2 = 1.5 rad / sec

top view of HORIZONTAL PLANE of motion for P

O P P R1 R₂ position 1 position 2 vP1 = 20 m/sec 36.87°

ω

O P e_θ eR F

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PROBLEM NO. 3

Part (d) – 4 points

Cart A and block B (having masses of M = 4 kg and m = 2 kg, respectively) are

connected by a spring of stiffness k = 300 N/m. The system is released from rest with the spring being compressed by 0.2 meters (position 1). Find the speed of cart A at position 2 when the spring is unstretched/uncompressed. Consider all surfaces to be smooth.

HINT: Consider both the energy and the linear momentum of system A + B as it moves.

F

!

x = 0 " lin. momentum in x # dir. conserved "

Mv_A1+ mv_B1= 0 = Mv_A2+ mv_B2 ! v_B2 = "2v_A2 No external forces do work. Therefore,

T₁+ V₁= T₂+ V₂ ! 0+ 1 2k"1 2 ₌ 1 2MvA2 2 ₊1 2mvB2 2 ₌ 1 2

(

M + 4m

)

vA2 2 _! v_A2 = k M + 4m "1= 300 4+ 8

( )

0.2 = 1 m / sec

Position 1

(at rest)

B k m M A A B

Position 2

(both A and B moving)

A

B

x y