Dynamics 4600:203 Homework 09 Due: April 04, 2008 Name:

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1 Dynamics 4600:03 Homework 09 Due: April 04, 008 Name: Please denote your answers clearly, i.e., box in, star, etc., and write neatly. There are no points for small, messy, unreadable work... please use lots of paper. Problem 1: Hibbeler, The 100lb block slides down the inclined plane for which the coefficient of kinetic friction is µ k = 0.5. If it is moving at 10ft/s when it reaches point A, determine the maximum deflection of the spring needed to momentarily arrest the motion. If δ measures the deflection of the spring, the total work done by the sliding friction, gravitational, and spring forces as the block moves from A is W 1 = mg (sinφ µ k cos φ) (d + δ) + k δ. The change in kinetic energy of the system is T = T T 1 = 0 m v. Therefore the equations of work-energy can be used to yield W 1 = mg (sin φ µ k cos φ) (d + δ) + k δ = m v = T. inally, we can rewrite this as a quadratic equation in terms of δ as [ ] ] k δ + mg (sinφ µ k cos φ) δ + [mg d (sin φ µ k cos φ) + mv = 0. With these parameter values, the maximum displacement is δ =.56ft. Problem : Hibbeler, 14 6 Cylinder A has a weight of 60lb and B has a weight of 10lb. Determine the distance A must descend from rest before it obtains a speed of 8ft/s. Also, what is the tension in the cord supporting A? Neglect the mass of the cord and pulleys. With the displacements of blocks A and B represented as x and y, the kinetic and potential energies are expressed as T = W A g ẋ + W B g ẏ, V = W A x + W B y. 1

2 These coordinates are constrained by the relation ẋ + ẏ = 0, so that y = x. Therefore the total energy can be expressed as E = T + V = W A + 4W B ẋ + (W A W B )x. g If the system is released from rest at x = 0, the initial energy is E 1 = 0, and the displacement may be solved in terms of the velocity as ( ) WA + 4W B ẋ x = W A W B g. With the given parameter values, we find that x =.48ft/s. With the tension acting on block A given as T ĵ, evaluating work-energy on block A for this system yields x 0 T dx W A x = W A g ẋ, so that Solving for T x 0 ( ) 6WA W B T dx = x. W A + 4W B T = 6W A W B W A + 4W B = 36lb. Problem 3: Hibbeler, Determine the smallest amount the spring at B must be compressed against the 0.50lb block so that when it is released from B it slides along the smooth surface and reaches point A. The velocity of the block when it reaches point A can be determined from conservation of energy. Evaluating the total energy of the block ( ) ( ) W k E = T + V = g v A + δ + W y, where δ is the compression of the spring and the velocity of the block is given as v A = ẋ + ẏ. The initial energy of the system is E 1 = k δ,

3 while the final energy, once the block makes it to point A at height h = 0.50ft = 6in is ( ) W E = g v A + W h. Therefore, with E 1 = E, the initial compression necessary to reach A with speed v A is ( ) W v δ = A k g + h. inally, the minimum compression, which leads to v A = 0, is simply W h δ = = 1.10in. k Problem 4: Hibbeler, Each of the two elastic rubber bands of the slingshot has an unstretched length of l = 00mm. If they are pulled back to the position shown d = 40mm and released from rest, determine the speed of the m = 5g pellet just after the rubber bands are become unstretched. Neglect the mass of the rubber bands and the change in elevation of the pellet while it is constrained by the rubber bands. Each rubber band has a stiffness of k = 50N/m. When each elastic band is pulled back a distance z, its total stretch s is given as s = z + h l, where h = 50mm is half the width of the slingshot and l = 00mm is its unstretched length. Therefore, the total energy of the system is ( ( ) m k ( m ( ) E = T + V = ż) + s = ż) + k z + h l. Therefore the initial energy is E 1 = k ( ) d + h l, while the final energy, with s = 0, is E = m ż Solving for ż ż = k m ( d + h l) =.86m/s. 3

4 Problem 5: Hibbeler, The Raptor is an outside loop roller coaster in which riders are belted into seats resembling ski-lift chairs. Determine the minimum speed v 0 at which the cars should coast down from the top of the hill, so that passengers can just make the loop without leaving contact with their seats. Neglect friction, the size of the car and passenger, and assume that each passenger and car has mass m. During the loop, the position of the chair can be described with the angular coordinate θ, so that its velocity and acceleration are ( v C = ρ θ ) ( ê θ, a C = ρ θ ) ( ê r + ρ θ ) ê θ. Identifying the speed of the chair as v = ρ θ, the velocity and acceleration become ( ) v v C = v ê θ, a C = ê r + v ê θ. ρ or the chair a free-body diagram is shown to the right. Note that the normal load N must be positive for the passengers to remain in their seats. N ê r m g ĵ With ĵ = sin θ ê r + cos θ ê θ, linear momentum balance becomes ( ( ) ) v = ( N mg sin θ) êr + ( mg cos θ) ê θ = m ê r + v ê θ = m a C. ρ Solving for the normal load N = mv ρ mg sinθ. The minimum speed required for the riders to remain in their seats at the top of the hill (θ = π/), so that N = 0, is v min = ρg. The total energy of the system is E = T + V = ( m v) + mg y, where y is the height of the chair above the ground. This energy must be conserved throughout the motion, so that in terms of the minimum velocity of the chair, the initial velocity of the ride must be v 0 = vmin + g (ρ h) = (5ρ h)g. 4

5 Problem 6: Hibbeler, The baseball has a horizontal speed of 35m/s when it is struck by the bat B. If it travels away at an angle of 60 from the horizontal and reaches a maximum height of 50m, measured from the height of the bat, determine the magnitude of the net impulse of the bat on the ball. The ball has a mass of 400g. Neglect the weight of the ball during the time the bat strikes the ball. The initial velocity of the ball is v B (t 1 ) = v 1 î = ( 35m/s) î, while the velocity of the ball after the impact is v B (t ) = v (cos θ î + sinθ ĵ), θ = 60, with v the unknown speed of the ball after the hit. However, with this velocity, the maximum height of the ball h is h = v sin θ. g inally, the impulse applied to the ball during the hit is I = ( m v B (t ) ) ( m v B (t 1 ) ) ( ) g cos θ = m h sin θ + v 1 = (1.N s) î + (1.5N s) ĵ, ( ) î + m g h ĵ, and the magnitude of the impulse is I = 4.7N s. Problem 7: Hibbeler, The m = 5kg block is moving downward at v 1 = m/s when it is h = 8m from the sandy surface. Determine the impulse of the sand on the block necessary to stop its motion. Neglect the distance the block dents into the sand and assume the block does not rebound. Neglect the weight of the block during the impact with the sand. Given the initial velocity of the block, its velocity v as it hits the sand is found from conservation of energy to be v = v1 + g h. Then, upon impact its change in linear momentum is L = m v B,3 m v B, = 0 m ( v ĵ), 5

6 and the total impulse acting on the block is I = L = mv ĵ = m v1 + g h ĵ = (63.4N s) ĵ. 6

THE WORK OF A FORCE, THE PRINCIPLE OF WORK AND ENERGY & SYSTEMS OF PARTICLES

THE WORK OF A FORCE, THE PRINCIPLE OF WORK AND ENERGY & SYSTEMS OF PARTICLES Today s Objectives: Students will be able to: 1. Calculate the work of a force. 2. Apply the principle of work and energy to