University of Alabama Department of Physics and Astronomy. PH 125 / LeClair Fall Exam III Solution

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1 University of Alabama Department of Physics and Astronomy PH 5 / LeClair Fall 07 Exam III Solution. A child throws a ball with an initial speed of 8.00 m/s at an anle of 40.0 above the horizontal. The ball leaves her hand.00 m above the round and experience neliible air resistance. (a How far from where the child is standin does the ball hit the round? (b How lon is the ball in fliht before it hits the round? Solution: Let the ball s startin heiht be y o, the initial velocity be v i, the launch anle be θ, and the horizontal distance traveled be d. If we put the oriin at round level where the child stands, such that the ball starts at coordinates (0, y o, we can use the y(x equation to find the distance traveled by findin the x coordinate at which y =0. y(x = 0 = y o + x tan θ x v i cos θ 0 = x + v i cos θ tan θx + v i y o cos θ ( 0 = x v i sin θ cos θx v i y o cos θ (3 ( Solvin the quadratic and simplifyin, x = = v i ( vi sin θ cos θ ± = v i cos θ ( sin θ cos θ ± cos θ 4vi sin θ cos θ + 8vi y o cos θ sin θ + y o ( sin θ ± sin θ + y o vi v i (4 (5 7.46,.03 m (6 Clearly the positive solution is the one we seek, so d=7.46 m. The time spent in the air is then easily found from the x component of the launch velocity: d = v x t = v i cos θt = t = d v i cos θ. s (7. A uniform solid sphere of mass M and radius R rotates with an anular speed ω about an axis throuh its center. A uniform solid cylinder of mass M, radius R, and lenth R rotates throuh an axis runnin throuh the central axis of the cylinder. What must be the anular speed of the cylinder so it will have the same rotational kinetic enery as the sphere? Solution: First: we don t need the lenth of the cylinder at all. All we need to do is equate the rotational kinetic enery of the two, with the sphere rotatin with ω s and the cylinder at ω c.

2 K r,cylinder = K r,sphere (8 I cωc = I sωs (9 Is 5 ω c = ω s = I ω s = ω s (0 c 5 3. Two blocks are connected by a strin that oes over an ideal pulley as shown in the fiure. Block A has a mass of 3.00 k and can slide over a rouh plane inclined 30.0 to the horizontal. The coefficient of kinetic friction between block A and the plane is Block B has a mass of.77 k. What is the acceleration of the blocks? m a θ m b Solution: We need free-body diarams for each mass. Let the x axis run up the ramp for mass A. We will assume that mass B falls and pulls mass A up the ramp, meanin the acceleration is in the x direction for mass A and downward for mass B. For mass A do we need to consider a friction force. Since the rope is presumably taut the entire time of interest, the acceleration is the same for both blocks. For the same reason, the tension applied to both blocks is the same. Newton s second law and eometry will suffice to find the acceleration if we nelect the pulley s rotational inertia. Alon the y direction for either mass, the forces must sum to zero, while alon the x direction, the forces must ive the acceleration for each mass. Fy = 0 Fx = ma x First consider mass A. The free body diaram above yields the followin, notin that the acceleration will be purely alon the x direction: Fy = n m a cos θ = 0 Fx = T f m a sin θ = m a a x From the first equation, we see n=m a cos θ, so f =µ k m a cos θ. For mass B, Fy = T m b = m b a = T = m b ( a

3 Combinin, T f m a sin θ = m b m b a µ k m a cos θ m a sin θ = m a a ( (m b + m a a = (m b µ k m a cos θ m a sin θ ( ( mb µ k m a cos θ m a sin θ a = 0.39 m/s (3 m a + m b 4. A record is dropped vertically onto a freely rotatin (undriven turntable. Frictional forces act to brin the record and turntable to a common anular speed. If the rotational inertia of the record is 0.54 times that of the turntable, what percentae of the initial kinetic enery is lost? Solution: The record starts at rest with the turntable rotatin with velocity ω i. Afterwards, both rotate toether with velocity ω f. Droppin the record on the turntable is essentially an inelastic rotational collision, so conservation of anular momentum will relate the two velocities. L i = L f (4 I t ω i = (I d + I t ω f (5 ( It ω f = ω i (6 I t + I d The ratio of final to initial kinetic enery is then readily found: The fraction lost is then K f K f = = K f = Approximately 35% of the initial kinetic enery is lost. (I t + I d ω f I tω i = I t I t + I d (7 I d = (8 I t + I d A strin is wrapped around a pulley with a radius of.0 cm and no appreciable friction in its axle. The pulley is initially not turnin. A constant force of 50 N is applied to the strin, which does not slip, causin the pulley to rotate and the strin to unwind. If the strin unwinds. m in 4.9 s, what is the moment of inertia of the pulley? Solution: The distance the strin travels x in a time t implies an acceleration a: x = at (9 Since the strin does not slip, the rotational acceleration of the pulley must match the linear acceleration of the strin divided by the radius of the pulley, α=a/r. A torque balance relates the acceleration to the force present:

4 τ = RF = Iα = I a R = a = R F I (0 x = at = R F I t ( I = R F t x 0.0 k m (

5 basics sphere Formula sheet = a free fall = 9.8 m/s V = 4 3 πr3 ax + bx + c = 0 = x = b ± b 4ac a D & D motion speed = v = v a x = lim t 0 v x t v x(t = v x,i + a xt v av r t dvx x(t = x i + v x,i t + axt v x,f = v x,i + ax x = d near earth s surface v = lim t 0 ( dx r t d r = d x launched from oriin, level round x y(x = (tan θ o x vo cos θ o max heiht = H = v i sin θ i Rane = R = v i sin θ i interactions U G = m x E mech = K + U a x a x = m m K = mv E mech = K + U = 0 E = W U sprin = k (x xo non-dissipative, closed P = de eneral P = Fext,x vx D const force Rotation: we use radians s = θr arclenth ω = dθ = vt r α = dω a t = αr tanential a r = v r = ω r radial v t = rω v r = 0 θ = ω i t + αt ω f = ω i + αt const α const α x = rθ v = rω a = rα no slippin I = i m i r i I z = I com + md L = r p = I ω K = Iω = L /I K = Iω f Iω i = W = P = dw = τω r dm = cmr I = mr point particle axis z parallel, dist d L = Iω = mvr τ = rf sin θ rf = r F = rf τ net = τ = I α = d L τ dθ K tot = K cm + K rot = mv cm + Iω work E mech = K + U = W not closed U sprin = k (x xo P = de P = F ext,x v x one dimension ( W = F x F constant foce D W = n x f W = (F ext,x x F n const nondiss., many particles, D x i F x(x dx nondiss. force, D (F s max = µsf n static F k = µ kf n kinetic W = F r F r f W = const non-diss force r i F( r d r variable nondiss force Moments of inertia of thins of mass M Object axis dimension I solid sphere central axis radius R 5 MR hollow sphere central axis radius R 3 MR solid disc/cylinder central axis radius R MR hoop central axis radius R MR point particle pivot point distance R to pivot MR rod center lenth L ML rod end lenth L 3 ML solid reular octahedron throuh vertices side a 0 ma Derived unit Symbol equivalent to newton N k m/s joule J k m /s = N m watt W J/s=m k/s3