Dynamics 4600:203 Homework 03 Due: February 08, 2008 Name:

Transcription

1 Dynamics 4600:03 Homework 03 Due: ebruary 08, 008 Name: Please denote your answers clearly, i.e., bo in, star, etc., and write neatly. There are no points for small, messy, unreadable work... please use lots of paper. Problem 1: Hibbeler, 1 96 A boy at O throws a ball in the air with a speed v 0 at an anle. If he then throws another ball at the same speed v 0 at an anle θ <, determine the time between the throws so the balls collide in mid air at B. see tetbook for fiure If the position of either ball is measured as r BO = +y, then the response of the ball durin fliht is ẍ = 0, ÿ =, ẋ = v 0 cos θ, ẏ = t + v 0 sin θ, = v 0 cos θt, y = t + v 0 sin θt. Eliminatin the time t from these equations, the vertical displacement y can be written in terms of the horizontal displacment as + v0 sin θ cos θ. y = v0 cos θ + sin θ cos θ = v 0 cos θ If the two balls collide there is some position,y so that y = v0 + v cos 0 sin cos, = v0 + v cos 0 sin θ cos θ. θ Notice that for a iven there is not just a sinle second anle θ for which the balls collide. Instead, for any other initial anle the balls will collide, but at different locations. Settin these equal yields v0 + v cos 0 sin cos = we solve for this horizontal displacement as = v 0 v0 + v cos 0 sinθ cos θ θ cos cos θ sin cos θ sin θ cos cos θ cos. Therefore the time required for each ball to travel this distance is t 1 = v 0 t = v 0 cos θ sin cos θ sin θ cos cos θ cos, cos sin cos θ sin θ cos cos θ cos. 1

2 inally, subtractin these, the time difference t = t 1 t becomes t = v 0 sin cos θ sin θ cos = v 0 cos θ + cos sin θ cos θ + cos Problem : Hibbeler, A car is travelin alon the circular curve havin a radius r = 400ft. At the instant shown, its anular rate of rotation is θ = 0.05rad/s, which is decreasin at the rate θ = 0.008rad/s. Determine the radial and transverse components of the car s velocity and acceleration at this instant and sketch these components on the curve. see tetbook for fiure With the position of the car measured in terms of polar coordinates r and θ, its position, velocity, and acceleration become r CO = r ê r, a C = r r θ ê r + v C = ṙ ê r + r θ ê θ, r θ + ṙ ω ê θ. or this motion the radius r is constant, so that ṙ = 0 and r = 0, and the velocity and acceleration reduce to v C = r θ ê θ, a C = r θ ê r + r θ ê θ. With the iven kinematic variables, these become v C = 10ft/s ê θ, a C = 0.5ft/s ê r + 3.ft/s ê θ. Problem 3: As illustrated in the fiure, the projectile, subject to only the influence of ravity, is launched with initial speed v 0, an anle θ 0 = 60, and a distance l = 10m from the ede. or what rane of initial speeds will the mass will reach the lower platform? v 0 θ 0 r P O = t + yt l l l

3 a As shown in the fiure, the position of the projectile P with respect to the firin point O is r P O = t + yt, so that t measures the horizontal distance travelled while yt measures the vertical displacement of P. The initial position of the projectile is r P O 0 = 0 + y0 = 0, while the initial velocity is v P 0 = ẋ0 + ẏ0 = v 0 cos θ + v 0 sinθ. The acceleration of the projectile is then determined to be a P = ẍ + ÿ. A suitable free-body diaram for the projectile is shown to the riht, where only the force due to ravity acts on P. Therefore the equations of motion can be written as = m a P, m = mẍ + ÿ. or takin components in the and directions m ẍ = 0, ÿ =. Usin the initial conditions 0 = 0, ẋ0 = v 0 cos θ, y0 = 0, ẏ0 = v 0 sin θ, these equations may be solved to yield t = v 0 cos θ t, yt = t + v 0 sin θ t. Notice that we can eliminate t from these two equations to directly relate and y as y = v 0 cos θ + sin θ v cos θ = 0 sinθ cos θ v0. cos θ At the minimum initial speed v 0 = v min the projectile just clears upper ede at,y = l,0, so that v 0 = min sin θ cos θ l l vmin. cos θ Solvin for v min vmin = l sin θ cos θ = l sinθ. Likewise, at the maimum initial speed the projectile reaches the lower ede, which is located at, y = 3 l/, l/, so that l v = ma sin θ cos θ 3 l 3 l vma cos, θ 3

4 and solvin for v ma 9 l v ma = 43 cos θ sin θ + cos θ = 9 l 1 + cos θ + 3sin θ. inally, the mass reaches the lower platform for v min < v 0 < v ma, or, with the values iven above 10.64m/s < v 0 < 11.94m/s. Problem 4: The particle P moves alon a spiral path, defined by: rθ = 1 π θ. If the anular position of the particle is defined as θt = ω t, find the position and acceleration of P with respect to the round after t = s if ω = 4rad/s. ê θ ê r P We measure the position of the particle P with respect to the oriin O in terms of polar coordinates: r P O = rt ê r. With this, the velocity and acceleration of P become: v P = ṙt ê r + rt θt ê θ, a P = rt rt θ t ê r + rt θt + ṙt θt ê θ. or the iven path: rt = ω t π, θt = ω t, ṙt = ω π, θt = ω, rt = 0, θt = 0, so that the position, velocity not required, and acceleration at t = s become: r P O = 8 π êr, v P = 4 π êr + 3 π êθ, a P = 18 π êr + 3 π êθ. inally, at this instant ê r and ê θ may be related to and as: ê r = cos θt ê r + sinθt ê θ, ê θ = sin θt ê r + cos θt ê θ, = ê r ê θ, = ê r ê θ. 4

5 Problem 5: Hibbeler, The blade on the horizontal-ais windmill is turnin with an anular velocity of ω 0 = rad/s. Determine the distance point P on the tip of the blade has traveled if the blade attains an anular velocity of ω = 5rad/s in 3 s. The anular acceleration is constant. Also, what is the manitude of the acceleration of this point when t = T = 3s? ê 1 r P ê O θ P r P = 15 ft We identify the directions ê 1 and ê, fied in the blade, toether with the anle θ that describes the inclination of the blade with respect to the round. With respect to the blade, the position of the tip is fied, and r P O = r p ê 1. With the anular velocity and anular acceleration identified as ω B/ = θ ˆk, α B/ = θ ˆk, the velocity and acceleration of P reduce to v P = ω B/ r P O = r θ ê, a P = α B/ r P O,+ω B/ ω B/ r P O = r θ ê 1 + r θê. The anular acceleration is constant, so that θ = α = constant. Therefore with T = 3s θt = α t + θ0, α = ω ω 0 T = 1rad/s, and the velocity of P becomes v P = r α t + ω 0 ê. The distance traveled by P is iven as the interal of speed over the duration of the motion. That is T α t T r ω + d = r α t + ω 0 dt = r 0 + ω ω0 0 t = T. 0 or these values, the total distance becomes d = ft. inally, the acceleration of P at this instant is a P = r θ ê 1 + r θê = 375ft/s ê ft/s ê. 5

6 The manitude of the acceleration is a P = ft/s. Problem 6: Hibbeler, The crankshaft AB is rotatin at a constant anular velocity of ω = 150 rad/s. Determine the velocity of the piston P at the instant θ = 30. ê ê 1 A ê 11 θ B ê 1 P The position of the piston P can be written as r P A = = l 1 ê 11 + l ê 1 = r BA + r P B. Therefore we can solve for and the inclination of link BP, defined by the anle φ, as l1 sin θ φ = arcsin, = l 1 cos θ ± l l 1 sinθ. l At θ = 30, these reduce to φ = 7.7 and = 0.9ft, takin the + solution in the above epression for. Then, the velocity of P, which is epressed as v P = ẋ, can also be written in terms of the intermediate link as v P = v B + ω / r P B, with v B = ω 1/ r BA. Therefore, this reduces to ẋ = θ ˆk l 1 ê 11 + φ ˆk l ê 1, ẋ = l 1 θ ê1 l φê. inally, ẋ can be solved as sinθ + φ ẋ = l 1 θ. cos φ inally, at this instant, the velocity of P reduces to v P = l 1 θ sinθ + φ = 18.5ft/s. cos φ 6

7 Problem 7: Hibbeler, The velocity of the slider block C is 4ft/s up the inclined roove. Determine the anular velocity of links AB and BC and the velocity of point B at the instant shown. see tetbook for fiure 7