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Unformatted text preview: Assignment 3 Solutions October 13th, 2010 1 Problem 3.4 We start with the standard kinematics equation for an object with an acceleration. In this case, acceleration is gravity. h = h0 + v0 t + 1 gt2 2 if we assume that the object is falling positive downward, and that h0 and v0 are zero and that the second ball is dropped at a time, dt , after the ﬁrst then we can track the position of the two objects as they fall: 1. h1 = 1 g (t + dt)2 2 2. h2 = 1 gt2 2 • subtracting h2 from h1 we get:
1 ⇒ 2 g (t2 + 2tdt + dt2 ) − 1 gt2 2 1 ⇒ 2 g (t2 − t2 + 2tdt + dt2 ) 1 ⇒ 2 g (2tdt + dt2 ) ∆h = h1 − h2 = 1 g (t + dt)2 − 1 gt2 2 2 the dt2 term stays constant while 2tdt increases with time, therefore as time increases, so does the distance, ∆h , between the two falling objects. answer: 1 g (2tdt + dt2 ) 2 2 Problem 3.8 Because we are only interested in crossing the river horizontally, we are only interested in the horizontal component of velocity. Because the river is ﬂowing perpendicular to our desired direction of travel, its velocity component is unimportant in our calculation. Though the boat will drift downstream with the river, as long as we continue to paddle directly to the bank, perpendicular to the ﬂow, none of the energy spent paddling will be wasted. answer: perpendicular to the current 1 3 Problem 3.10 Because the object is projected oﬀ the roof with only horizontal velocity, we proceed initially as if the object was dropped in order to ﬁnd the amount of time it took for the object to fall, we start with our standard ballistics relationship for an object with an acceleration, as before acceleration is gravity: h = h0 + v0 t + 1 gt2 2 • if we assum that the object starts with the roof as the zero point, and it is falling positive downward a distance, h, we get: h h = 1 gt2 , solving for t we get t = 2g 2 • so if we were to subsitute h with our two building heights H and 2H , we get: H H h = H −→ TH = 2g and h = 2H −→ T2H = 4g • we can rewrite the 2H case, with a little math trickery as √H √ T2H = 2 2g = 2TH √ 2TH answer: T2H = 2 4 Problem 3.9 Because the object rolls oﬀ the table, it has no initial velocity in the y direction, Vy 1. Find the time of ﬂight of the ball • h = h0 + v0 t + 1 gt2 2 answer: 0.355s • if we assum that the object starts with the roof as the zero point, and it is falling positive downward a distance, h, we get: 55m h • h = 1 gt2 , solving for t we get t = 2g −→ t = 2(0.m/s2) = 0.35s 2 9 .8 2. Find the initial magnitude of velocity • x = x0 + v0 t + 1 gt2 2 • we assume that initial position, x0 , is zero and acceleration, a, is zero • x = v0 t −→ 1.3m = v0 (0.35s) solve for v0 : answer: v0 = 3.88m/s 3. Find the magnitude of velocity just before the ball strikes the ﬂoor • vf y = v0y + gt • v0y = 0 so vf y = gt −→ vf y = 9.8m/s2 (0.35s) = 3.28m/s • vf x = 3.88 from above 3 • mag (v ) or [v ] = 5.08m/s] 2 2 vx + vy −→ (3.88m/s)2 + (3.28m/s)2 = answer: [v ] = 5.08m/s] 4. Find the direction of velocity just before the ball strikes the ﬂoor tan θ =
opp adj v 3.28m/sy 3.88m/s )
y θ = tan−1 ( vx ) −→ tan−1 ( = 40.2o answer: 40.2o below horizontal 5 Problem 3.20 When y0 = yf it is best to break the problem up into parts, ﬁrst ﬁnding the time it takes to reach the apex of the trajectory, where we know that vy = 0, then once we have found the elevation of the peak, we can calculate the rest of the problem as in 3.9 and 3.10. before proceeding we should recognize that the velocity needs to be broken up into its components, vx and vy : vx = v0 cos θ vy = v0 sin θ 1. Calculate the maximum height above the roof reached by the rock h = h0 + v0 t + 1 gt2 2 vay = v0y − gt • solving for t we get t = • t=
30m/s·sin 34 9 .8
0 equation 1a equation 1b
v0y g and plugging this into equation 1a yields = 1.71s 4 • h = v0 v0y g answer: h = 0 1 (30m/s·sin 34 ) 2 9 .8 − 1g 2 v0y g 2
2 =⇒ 2 1 v0y 2g = 14.35m 2. Calculate the magnitude of the velocity of the rock just before it strikes the ground • Now that we have calculated the height and time to reach the apex of the trajectory we can proceed with the rest of the problem using those quantities, as if the rock were sliding oﬀ apex with an initial velocity as in the previous problem. If we take the apex to be at h = h0 + hb and assume that we are traveling downward to h = 0, we get • h = h0 + hb = 1 gt2 2 + 4 • if we solve for t we get t = 2(h0g hb ) = 2(19.0+14.2 m) = 2.61s 9.8m/s • now that we have the time it takes the rock to fall, we can calculate how fast it is falling at in the y direction before it hits the ground • vy = gt = 9.8m/s2 ∗ 2.61s = 25.5s • now we know that there is no acceleration in the x direction, so calculation vx is trivial vx = v0 cos θ = 30m/s · cos 34o = 24.5m/s 2 2 [v ] = vx + vy −→ (24.5m/s)2 + (25.5m/s)2 = 35.6m/s answer: [v ] = 35.6m/s 3. Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground. • Now that we have the time it takes to reach the apex and the time it takes to fall to the ground, ﬁnding the horizontal distance is as simple as • xtotal = vx ∗ ttotal = vx ∗ (t0A + tAB ) = 24.5m/s ∗ (1.71s + 2.61s) answer: xtotal = 105.9 6 Problem 3.23 5 known quantities: • v0 = 23.0m/s • ttotal = 3.20s • dtotal = 37.0m 1. Find the angle of elevation of the hose. (a) there is no acceleration in the x direction so we have • x = v 0x t • as before we have: vx = v0 cos θ vy = v0 sin θ • therefore x = v0 t · cosθ 37m • solving for θ = cos−1 vdtotal −→ θ = cos−1 23.0m/s·3.2s 0 ttotal answer: θ = 59.8o 2. Find the speed of the water at the highest point in its trajectory. • since we know that at the apex of the trajectory, all speed is in the xdirection and since gravity is the only force then • vhp = vx = v0 cos θ −→ vhp = 23m/s · cos (59.8o ) answer: vhp = 11.6m/s 3. Find the acceleration of the water at the highest point in its trajectory. • Because gravity is the only force acting on the system, then the acceleration is answer: a = 9.8m/s2 4. How high above the ground does the water strike the building? • vy = v0 − gt • vy = 23m/s · sin(59.8o ) − 9.8m/s2 · 3.2s answer: vy = 11.43m/s downward 5. How fast is it moving just before it hits the building? • vx = v0 cos θ −→ vhp = 23m/s · cos (59.8o ) = 11.6m/s to the right 2 2 [v ] = vx + vy −→ (11.6m/s)2 + (11.43m/s)2 = 16.29m/s answer: [v ] = 16.29m/s 6 7 Problem 3.29 This problem has two components, derivation of the range formula, and maximizing the range of the trajectory as a function of a variable theta Derivation of the range formula: y = v0 t − 1 gt2 = v0 t · sinθ − 1 gt2 equation 1a 2 2 x = v0x t = v0 t · cosθ equation 1b • solve for t = y=
v0 ·sinθ v0 ·cosθ x v0y cosθ because we are looking the trajectories maximum range, we set y = 0 and solve for x which would correspond to two points on the trajectory, the beggining or when the cannonball falls. 2 θ – sinθ · x = 1 g v0yx θ cos 2 cos
x – 2sinθ · cosθ · x = g v2
2 0y · x − 1g 2 and plug into equation 1a 2
x v0y cosθ – sin2θ = g vx 2 – R=x= 0y 2 v0y g sin2θ 1. f a certain cannon can shoot a projectile a distance of 260m when it is aimed at 66.0◦ above the horizontal, what is the maximum range the cannon could achieve with the same projectile? • using the range formula we can solve for the intial velocity of the cannon – R=
2 v0y g sin2θ • Here’s the tricky part, in order to ﬁnd the maximum range, we also have to ﬁnd the optimal angle. From calculus, we know that this is done by minimizing the derivative, here we’re going to take the derivative of R with respect to θ. –
dR dθ – =⇒ v0y = R sin2θ = 260m sin2·66.0o = 58.55m/s =2 2 v0y g cos2θ 7 • Now we’ll set – – – – –
2 v0y dR dθ = 0 and solve for theta 2 g cos2θ = 0 cos2θ = 0 2θ = cos−1 (0) 2θ = 90o θ = 45o • Now we can solve for maximum range using the range equation with θ = 45.0◦ . • =⇒ Rmax =
2 v0y g sin2θmax = 58.55m/s 9.8m/s2 sin2 · 45o 2. At what angle should it be aimed to do this? answer: θmax = 45o answer: =⇒ Rmax = 349.9m 8 Problem 3.31 Equation for rotational motion: 2 arad = vr 1. At what speed should you swing it so its acceleration will be 9.8m/s2 ? m = 2.2kg l = 0.80m √ v = arad ∗ R = 9.8m/s2 ∗ 0.8m answer: v = 2.8m/s 9 Problem 3.37 knowns: dθ dt = 3.20rotations/s r = 0.07m 1. What is the radial acceleration (inm/s2 ) of the endolymph ﬂuid? • because we’re rotating, the tangential velocity is going to be equal to the number of circumferances that rotate about that circle per cycle – vt = – arad · 2π · r 2 2 2 ( d θ ·2 π ·r ) 2 = vr = dt r = 4 π dθ r =⇒ 4 (π 3.20rot/s) 0.07m dt
dθ dt answer: arad = 28.3m/s2 2. What is the radial acceleration (in g’s) of the endolymph ﬂuid? answer: arad /g = 28.3m/s2 /9.8m/s2 = 2.88g s 8 ...
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This note was uploaded on 01/17/2011 for the course PHYS 6a taught by Professor Stanek during the Fall '07 term at UCSB.
 Fall '07
 STANEK
 Acceleration, Gravity

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