+ 1 2 mv 2. Since no forces act on the system in the x-direction, linear momentum in x-direction is conserved: (2) 0 = mv A2. + Rω 2.

Transcription

1 ME 74 Spring 018 Final Exaination Proble 1 Given: hoogeneous dis of ass and outer radius R is able to roll without slipping on the curved upper surface of a cart. art (of ass ) is able to ove along a sooth, horizontal surface. The syste is released fro rest at Position 1. t Position, the center of the dis has dropped through a vertical distance of H, and the dis is rolling on a horizontal portion of the upper surface of cart. Find: For this proble: (a) Draw a single free body diagra (FBD) of the syste ade up of the dis and the cart. (b) Explain why echanical energy is conserved for the syste shown in your FBD. (c) Explain why linear oentu in the x-direction is conserved for the syste shown in your FBD. (d) Deterine the angular velocity of the dis at Position. Write your answer as a vector. Posi%on 1 R H Posi%on y sooth x datu 1. FBD - above. Kinetics Since no non-conservative wor is done on the syste, energy is conserved: (1) T 1 +V 1 = T +V 0 + gh = 1 v + 1 v + 1 I ω + 0 ; I = 1 R Since no forces act on the syste in the x-direction, linear oentu in x-direction is conserved: () 0 = v + v v = v 3. Kineatics (3) v = v î = v + ω r / = v î + ω 4. Solve obining (1)-(3): gh = ( ) 1 v R ω = R ω ( ) ( Rĵ) = v + Rω R ω = 1 R ω ( )î v = R ω ω = gh R ˆ

2 ME 74 Spring 018 Final Exaination Proble Given: hoogeneous dis of ass and outer radius R is able to roll without slipping on a rough horizontal surface. bar (of ass ) is in a no-slip contact with the top of the dis with the other end of being supported by a sooth roller. spring of sti ness is connected between the right end of bar and ground. second spring (of sti ness ) is connected between the center of the dis and a ovable base B. Base B is given a prescribed otion of x B (t) =b sin t. Let x represent the otion of the bar, where x = x B = 0 when the springs are unstretched. Find: For this proble: (a) Draw individual free body diagras of the dis and of the bar. (b) Derive the di erential equation of otion (EM) for the syste in ters of the coordinate x. (c) What is the natural frequency n of the syste? (d) Deterine the particular solution x P (t) of your EM. (e) Based on your particular solution above, is the otion of the bar in phase or 180 out of phase with the otion of the base B when = p /? x + x B (t) B R sooth f x 1. FBDs - shown. Newton-Euler Bar: = x + f = x Dis: F x Therefore: M = ( Rθ x B ) R f ( R) = I θ ; I = I + R = 1 R + R = 3 R ( ) (1) f = x + x = 3 4 R θ Rθ x B 3. Kineatics Since is the I for the dis: () x = R θ θ = x / R ND θ = x / R 4. EM obining (1) and (): x + x = 3 4 R x R R x R x B 11 8 x + 3 x = b sinωt Divide EM by the coefficient of x gives: x x = 8 11 b sinωt x + ω nx = F 0 sinωt ; F 0 = 8 11 b with: ω n = 1 11 ( Rθ x B ) f f θ +

3 Particular solution x P t x P t ( ) = cosωt + BsinΩt ( ) = Ω cosωt Ω BsinΩt Substituting in the EM gives: Ω ( + ω n ) cosωt + Ω + ω n cosωt : = 0 sinωt : Ω + ω n ( ) BsinΩt = F 0 sinωt ( ) B = F 0 B = 8 / 11 ( )( / ) Ω + ω n Since Ω = / < ω n B > 0 particular solution response is IN PHSE with the base otion. b

4 ME 74 Spring 018 Final Exaination Proble 3 Given: The following equation of otion (EM) has been derived for a single-degree-of-freedo syste: 6ẍ ẋ + 16x = 0. Find: ircle the figure below that ost accurately represents the response that is described by the above EM. response, x(t) tie, t tie, t response, x(t) response, x(t) tie, t EM: x 1 x + 36x = 0 3 ζω n = 1/ 3 ζ < 0 x( t) = e ζ ω n t ( cosω d t + S sinω d t) exponentially GRWING oscillations

5 ME 74 Spring 018 Final Exaination Proble 3B Given: onsider Systes and B shown below. Syste is ade up of a spring and bloc with the bloc oving in pure translation along a sooth horizontal surface. Syste B is ade up of a spring and a hoogeneous dis of ass and outer radius R, with the center of the dis at and the dis rolling without slipping on a horizontal surface. Each syste has the sae ass and sae spring sti ness. Let n and nb represent the natural frequencies of Systes and B, respectively. Find: ircle the answer below that ost accurately represents the natural frequencies for the two systes: (a) n < nb (b) n = nb (c) n > nb x x sooth R Syste Syste B x x f Syste : F x = x = x x + x = 0 x + x = 0 ω n = Syste B: M = x( R) = I θ = Therefore, ω n > ω nb. 3 x R R 3 x + x = 0 x + 3 x = 0 ω nb = 3

6 ME 74 Spring 018 Final Exaination Proble 3 Given: onsider the free response for a daped, single-df syste having the following di erential equation of otion: ẍ + cẋ + x = 0. This syste is nown to have a daping ratio of = 0.1 and undaped natural frequency of n = 10 rad/s. Find: What is the new value for the daping ratio : (a) the original value of is doubled, the original value of is doubled and the value of c is unchanged? (b) The original value of is doubled, the original value of c is doubled and the value of is unchanged? x + cx + x = 0 x + c x + x = 0 ω n = & ζ = c ω n = c a) ζ new = ( )( ) = 1 ζ old c b) ζ new = = ζ ( old ) c

7 ME 74 Spring 018 Final Exaination Proble 3D Given: cable is wrapped around the inner radius of a stepped dru, with the other end of the cable attached to ground, as shown in the figure. second cable is wrapped around the outer radius of the dru, with a force F applied to the free end of that cable. The dru is supported by a horizontal surface on which the dru slips at contact point as the dru oves. Neither cable slips on the dru as the syste oves. Let µ s and µ be the coe cients of static and inetic friction, respectively, between the dru and the supporting surface at. Find: ircle the answer below that ost accurately describes the friction force f acting on the dru at the contact point. (a) f = µ s g (to the left) (b) f = µ g (to the left) (c) f =0 (d) f = µ g (to the right) (e) f = µ s g (to the right) F taut cable slip R R Note that is the I for the dru. Therefore, as the force F is applied, the dru will rotate W about. s a result, the contact point will ove to the LEFT. Since friction opposed the sliding otion of, the sliding friction force f = µ g will act to the RIGHT.

8 ME 74 Spring 018 Final Exaination Proble 3E Given: particle P of ass is free to slide on a sooth bar of negligible ass. The bar is free to rotate in a horizontal plane about a vertical axis passing through end of the bar. spring of sti ness and unstretched length R 0 is connected between P and. The spring is copressed to half of its unstretched length and released when the bar has a rotational speed of 1. fter release, P reaches a position when the spring is unstretched. t this position, the rotational speed of the bar is. Find: Suppose now the experient is repeated except the sti ness of the spring is doubled to a value of. s a result of this change, the value of is now: (a) decreased. (b) unchanged. (c) increased. rotational speed of the bar is. P R 0 / 1 P HRIZNTL PLNE R 0 For the syste of the bar and particle P, M = 0. Therefore: H 1 = H r P/ v P1 = r P/ v P R 0 êr R 0 R 0 ω 1êθ = R0êR ( ) ( "Rê R + R 0 ω ê θ ) ( ) 4 ω 1 ˆ = R 0 ω ˆ ω = 1 4 ω 1 INDEPENDENT of

9 ME 74 Spring 018 Final Exaination Proble 3E Given: ircraft B is traveling along a straight path with a speed of v B. ircraft is traveling along a circular path with a speed of v. Find: ircle the answer below that ost accurately describes the observed velocities of and B: (a) ~v /B = ~v ~v B is the velocity of as seen by the pilot of B. (b) ~v B/ = ~v B ~v is the velocity of B as seen by the pilot of. (c) Both (a) and (b). (d) Neither (a) nor (b). v B v B B v vb ( v / B ) = v rel v B ω r / B = v / B ω r / B = vel. of as seen by B. Here ω = ang. vel. of B = 0. Therefore: ( v / B ) = v rel / B ( v B/ ) = v rel B v ω r B/ = v B/ ω r B/ = vel. of B as seen by. Here ω = ang. vel. of 0. Therefore: ( v B/ ) v rel B/