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

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1 RLEM N. 1 Given: Find: vehicle having a mass of 500 kg is traveling on a banked track on a path with a constant radius of R = 1000 meters. t 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/sec (due to braking). For this instant, determine the magnitude of the TTL 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 FD for your analysis. e t R k R v e n θ = e n T VIEW FD s e t k mg e n N e n f turn f brake T VIEW

2 Examination No. 1 FD: previous page Newton: (1)! F t = " f brake = ma t ()! F n = " f turn cos# + N sin# = ma n (3)! F z = f turn sin# + N cos# " mg = ma z $ ( ) f turn sin# + N cos# = m g + a z Multiply () by cosθ, (3) by sinθ and subtract: f turn ( cos! + sin!) = "ma n cos! + m( g + a z )sin! # (4) f turn = m $ %"a n cos! + ( g + a z )sin! & ' Kinematics: (5) a t = dv =!4 m / sec dt (6) a n = v R (7) a z = 0 Solve: Combining (1), (4)-(7): f brake =!ma t =! 500 ( )!4 ( ) = 000 N # f turn = m! v & % cos" + gsin" ( $ % R '( = 500 # % $ % Therefore, f total = f brake + f turn ( )! = = 787 N ( ) ( 0.6 ) & ( '( = 1941 N

3 RLEM N. Given: Find: articles and (having masses of m = m = 10 kg) are interconnected by the cable-pulley system shown in the figure. oth particles are constrained to vertical motion with particle able to slide on a smooth vertical rod. The system is released at s = 0 with traveling downward with a speed of 5 m/sec. ssume the pulleys to be small, massless and frictionless. You are asked here to find the speed of particle when has reached the position of s = m. In your solution clearly indicate the following of the four-step solution method: 1. FD: Complete the free body diagram of the system of, and the cable shown below right. Identify any nonconservative forces that do work on the system.. Work-energy equation: Clearly indicate the gravitational datum line(s) used. 3. Kinematics: Here you need to relate s to s as well as v to v for the two positions. 4. Solve 1.5 m y datum for s s x g smooth rod datum for N m g m g LESE STRT YUR WRK N THE NEXT GE.

4 FD: shown on previous page. ll forces doing work are conservative and will be included in the potential energy. Work-energy: T 1 = 1 mv mv 1 V 1 = 0 T 1 = 1 mv + 1 mv V =!mgs + mgh ( ) = 0 nc U 1" Therefore, ( nc) T 1 + V 1 + U 1! = T + V " v 1 + v 1 = v Kinematics s + s = const.! + v + g (#s + h )!s + s!s s = 0! v = 1 s s v lso,!s = 1 # s " 1.5& $% '( = h t position 1: v 1 = v 1 = 0 t position : v = 1 Solve: v = v 1 + g( s! h ) =.5 v = 0.4 v and h = 1 ( 5) + ( ) ( 9.806) (! 0.5) (.5! 1.5 ) = 0.5 m = 6.85 m / sec

5 RLEM N. 3 art (a) 6 points article is attached to rigid bar with bar pinned to ground at. article strikes the stationary particle with a speed of v 1 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 HRIZNTL plane. Consider the following 1 statements about momentum and energy during impact and indicate if each statement is TRUE or FLSE RIGID bar v 1 = 0 HINT: Complete the FD of each of the three systems shown below prior to responding. v 1 For System + linear momentum in the n-direction is conserved: TRUE or FLSE linear momentum in the t-direction is conserved: TRUE or FLSE angular momentum about point is conserved: TRUE or FLSE mechanical energy is conserved: TRUE or FLSE t n System + For System linear momentum in the n-direction is conserved: TRUE or FLSE linear momentum in the t-direction is conserved: TRUE or FLSE angular momentum about point is conserved: TRUE or FLSE mechanical energy is conserved: TRUE or FLSE t n System For System linear momentum in the n-direction is conserved: TRUE or FLSE linear momentum in the t-direction is conserved: TRUE or FLSE angular momentum about point is conserved: TRUE or FLSE mechanical energy is conserved: TRUE or FLSE n t System

6 RLEM N. 3 art (b) 6 points n arm rotates about the fixed Z-axis with a rate of Ω. 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. t the instant shown, the xyz and XYZ axes are aligned. lso at this position, Ω = 3 rad/sec with Ω increasing at a rate of!! = rad / sec. 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 + p d j dt =!" K + "0 + ( 0) j + p # $ j = k + 4 = %1i + k ( ) (( ) $ j) ( ) 3k + 4 j ( ) rad / sec

7 RLEM N. 3 art (c) 4 points article (having a mass of m) is able to slide on a smooth HRIZNTL surface. n extensible cord (having a stiffness of 50 N/m and unstretched length of meters) is attached between and a fixed point in the plane of motion for. t position 1, is released with a velocity as shown below with R 1 = meters. ssuming that the cord remains taut for all time, find the angular speed ω of the cord about point when is at position where R = 4 meters. HINT: consider the angular momentum of as it moves about the fixed point. position v 1 = 0 m/sec R ω R 1 position 1 top view of HRIZNTL LNE of motion for! M = " H = H 1 " m r /! v = m r /! v 1 ( R e R )! (!R e R + R " e # ) = ( R 1 e R )! ( v 1 cos#e R + v 1 sin#e # ) R " k = R 1 v 1 sin#k $ e θ e R F! = R 1v 1 sin" R = ( )( 0) ( 0.6) ( 4) = 1.5 rad / sec

8 RLEM N. 3 art (d) 4 points Cart and block (having masses of M = 4 kg and m = 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. meters (position 1). Find the speed of cart at position when the spring is unstretched/uncompressed. Consider all surfaces to be smooth. HINT: Consider both the energy and the linear momentum of system + as it moves. M k m osition 1 (at rest) osition (both and moving)! F x = 0 " lin. momentum in x # dir. conserved " Mv 1 + mv 1 = 0 = Mv + mv! v = "v No external forces do work. Therefore, T 1 + V 1 = T + V! y k" 1 = 1 Mv + 1 mv = 1 ( M + 4m)v! x v = k M + 4m " 1 = 300 ( ) = 1 m / sec

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